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##### Logarithm

Logarithm Properties
There is a close relationship between logarithms and  The logarithm of a number is defined as the power or index to which a given base must be raised to obtain the number.

If we have $\displaystyle {{a}^{x}}=M$ where a  and M are greater than zero and a≠1 then we can write this in a logarithm form.

$\displaystyle {{\log }_{a}}M=x$

Since logarithms are closely related with then also their properties are similar.

### Logarithm of a Product

With when you multiply two numbers with the same base you add the exponents meanwhile with logarithms the logarithm of a product is the sum of the logarithms.

The logarithm of a product is the sum of logarithms.

$\displaystyle {{\log }_{a}}(M~N)={{\log }_{a}}M+{{\log }_{a}}N$

Example 1: Use the product property to evaluate.

a) $\displaystyle {{\log }_{3}}(3\cdot 9)$

We can evaluate it directly by multiplying the product and finding the value

$\displaystyle {{\log }_{3}}(3\cdot 9)=$

$\displaystyle {{\log }_{3}}(27)=x$

$\displaystyle {{3}^{x}}=27$

$\displaystyle {{3}^{x}}={{3}^{3}}$

$\displaystyle x=3$

But our exercise is asking us to use the product property so the solution would be

$\displaystyle {{\log }_{3}}(3\cdot 9)=$

$\displaystyle {{\log }_{3}}3+{{\log }_{3}}9=$

Now we evaluate each of them separately

$\displaystyle {{\log }_{3}}3={{x}_{1}}$

$\displaystyle {{3}^{{{{x}_{1}}}}}=3$

$\displaystyle {{3}^{{{{x}_{1}}}}}={{3}^{1}}$

$\displaystyle {{x}_{1}}=1$ and $\displaystyle {{\log }_{3}}9={{x}_{2}}$

$\displaystyle {{3}^{{{{x}_{2}}}}}=9$

$\displaystyle {{3}^{{{{x}_{2}}}}}={{3}^{2}}$

$\displaystyle {{x}_{2}}=2$

So $\displaystyle {{\log }_{3}}(3\cdot 9)=$

$\displaystyle {{\log }_{3}}3+{{\log }_{3}}9=$

$\displaystyle 1+2=3$

b) $\displaystyle {{\log }_{5}}(5\cdot 25)$

We can evaluate it directly by multiplying the product and finding the value

$\displaystyle {{\log }_{5}}(5\cdot 25)=$

$\displaystyle {{\log }_{5}}(125)=x$

$\displaystyle {{5}^{x}}=125$

$\displaystyle {{5}^{x}}={{5}^{3}}$

$\displaystyle x=3$

But our exercise is asking us to use the product property so the solution would be

$\displaystyle {{\log }_{5}}(5\cdot 25)=$

$\displaystyle {{\log }_{5}}5+{{\log }_{5}}25=$

Now we evaluate each of them separately

$\displaystyle {{\log }_{5}}5={{x}_{1}}$

$\displaystyle {{5}^{{{{x}_{1}}}}}=5$

$\displaystyle {{5}^{{{{x}_{1}}}}}={{5}^{1}}$

$\displaystyle {{x}_{1}}=1$ and $\displaystyle {{\log }_{5}}25={{x}_{2}}$

$\displaystyle {{5}^{{{{x}_{2}}}}}=25$

$\displaystyle {{5}^{{{{x}_{2}}}}}={{5}^{2}}$

$\displaystyle {{x}_{2}}=2$

So $\displaystyle {{\log }_{5}}(5\cdot 25)=$

$\displaystyle {{\log }_{5}}5+{{\log }_{5}}25=$

$\displaystyle 1+2=3$

Example 2: Evaluate the sum of logarithms.

a) $\displaystyle {{\log }_{6}}2+{{\log }_{6}}3$

Finding each of them separately is difficult

$\displaystyle lo{{g}_{6}}3={{x}_{2}}$

$\displaystyle {{6}^{{{{x}_{2}}}}}=3$

And $\displaystyle {{\log }_{6}}2={{x}_{1}}$

$\displaystyle {{6}^{{{{x}_{1}}}}}=2$

Based on the rule we know that the sum of the logarithms is equal with the product of logarithm.

$\displaystyle {{\log }_{6}}2+{{\log }_{6}}3=$

$\displaystyle {{\log }_{6}}(2\cdot 3)=$

$\displaystyle {{\log }_{6}}(6)=x$

Now the logarithm form we write it on an exponential form to find the answer

$\displaystyle {{6}^{x}}=6$

$\displaystyle {{6}^{x}}={{6}^{1}}$

$\displaystyle x=1$

So $\displaystyle {{\log }_{6}}2+lo{{g}_{6}}3=1$

b) $\displaystyle {{\log }_{2}}3+{{\log }_{2}}\frac{4}{3}$

Finding each of them separately is difficult

$\displaystyle {{\log }_{2}}3={{x}_{1}}$

$\displaystyle {{2}^{{{{x}_{1}}}}}=3$ and $\displaystyle {{\log }_{2}}\frac{4}{3}={{x}_{2}}$

$\displaystyle {{2}^{{{{x}_{2}}}}}=\frac{4}{3}$

Based on the rule we know that the sum of the logarithms is equal with the product of logarithm.

$\displaystyle {{\log }_{2}}(3\cdot \frac{4}{3})=$

$\displaystyle {{\log }_{2}}3+{{\log }_{2}}\frac{4}{3}=$

$\displaystyle {{\log }_{2}}4=x$

Now the logarithm form we write it on an exponential form to find the answer

$\displaystyle {{2}^{x}}=4$

$\displaystyle {{2}^{x}}={{2}^{2}}$

$\displaystyle x=2$

So $\displaystyle {{\log }_{2}}3+{{\log }_{2}}\frac{4}{3}=2$

Note! If the product has many factors, you just add the individual logarithms.

If $\displaystyle {{x}_{1}}>0,{{x}_{2}}>0,…….,{{x}_{n}}>0$ then

$\displaystyle {{\log }_{a}}({{x}_{1}}\cdot {{x}_{2}}…….{{x}_{n}})={{\log }_{a}}{{x}_{1}}+{{\log }_{a}}{{x}_{2}}+………+{{\log }_{a}}{{x}_{n}}$

Logarithm of a Quotient

With exponents when you divide two numbers with the same base you subtract the exponents meanwhile with logarithms the logarithm of a quotient is the difference of the logarithms.

The logarithm of a quotient is the difference of the logarithms.

$\displaystyle {{\log }_{a}}\left( {\frac{M}{N}} \right)={{\log }_{a}}M-{{\log }_{a}}N$

Example 3: Use the quotient property to evaluate.

a) $\displaystyle {{\log }_{2}}\left( {\frac{{32}}{8}} \right)$

We can evaluate it directly by dividing the quotient and finding the value

$\displaystyle {{\log }_{2}}\left( {\frac{{32}}{8}} \right)=$

$\displaystyle {{\log }_{2}}(4)=x$

$\displaystyle {{2}^{x}}=4$

$\displaystyle {{2}^{x}}={{2}^{2}}$

$\displaystyle x=2$

But our exercise is asking us to use the quotient property so the solution would be

$\displaystyle {{\log }_{2}}\left( {\frac{{32}}{8}} \right)=$

$\displaystyle {{\log }_{2}}32-{{\log }_{2}}8=$

Now we evaluate each of them separately

$\displaystyle {{\log }_{2}}32={{x}_{1}}$

$\displaystyle {{2}^{{{{x}_{1}}}}}=32$

$\displaystyle {{\log }_{2}}32={{x}_{1}}$

$\displaystyle {{2}^{{{{x}_{1}}}}}=32$

$\displaystyle {{2}^{{{{x}_{1}}}}}={{2}^{5}}$

$\displaystyle {{x}_{1}}=5$ and $\displaystyle {{\log }_{2}}8={{x}_{2}}$

$\displaystyle {{2}^{{{{x}_{2}}}}}=8$

$\displaystyle {{2}^{{{{x}_{2}}}}}={{2}^{3}}$

$\displaystyle {{x}_{2}}=3$

So $\displaystyle {{\log }_{2}}\left( {\frac{{32}}{8}} \right)=$

$\displaystyle {{\log }_{2}}32-{{\log }_{2}}8=$

$\displaystyle 5-3=2$

b) $\displaystyle {{\log }_{7}}\left( {\frac{{49}}{{2401}}} \right)$

We can evaluate it directly by dividing the quotient and finding the value

$\displaystyle {{\log }_{7}}\left( {\frac{{49}}{{2401}}} \right)=$

$\displaystyle {{\log }_{7}}\left( {\frac{1}{{49}}} \right)=x$

$\displaystyle {{7}^{x}}=\frac{1}{{49}}$

$\displaystyle {{7}^{x}}={{49}^{{-1}}}$

$\displaystyle {{7}^{x}}={{7}^{{-2}}}$

$\displaystyle x=-2$

But our exercise is asking us to use the quotient property so the solution would be

$\displaystyle {{\log }_{7}}\left( {\frac{{49}}{{2401}}} \right)=$

$\displaystyle {{\log }_{7}}49-{{\log }_{7}}2401=$

Now we evaluate each of them separately

$\displaystyle {{\log }_{7}}49={{x}_{1}}$

$\displaystyle {{7}^{{{{x}_{1}}}}}=49$

$\displaystyle {{7}^{{{{x}_{1}}}}}={{7}^{2}}$

$\displaystyle {{x}_{1}}=2$ and $\displaystyle {{\log }_{7}}2401={{x}_{2}}$

$\displaystyle {{7}^{{{{x}_{2}}}}}=2401$

$\displaystyle {{7}^{{{{x}_{2}}}}}={{7}^{4}}$

$\displaystyle {{x}_{2}}=4$

So $\displaystyle {{\log }_{7}}\left( {\frac{{49}}{{2401}}} \right)=$

$\displaystyle {{\log }_{7}}49-{{\log }_{7}}2401=$

$\displaystyle 2-4=-2$

Example 4: Evaluate the difference of logarithms.

a) $\displaystyle {{\log }_{3}}7-{{\log }_{3}}\frac{7}{9}$

Finding each of them separately is difficult

$\displaystyle {{\log }_{3}}7={{x}_{1}}$

$\displaystyle {{3}^{{{{x}_{1}}}}}=7$

And $\displaystyle {{\log }_{3}}\frac{7}{9}={{x}_{2}}$

$\displaystyle {{3}^{{{{x}_{2}}}}}=\frac{7}{9}$

Based on the rule we know that the difference of the logarithms is equal with the quotient of logarithms

$\displaystyle {{\log }_{3}}7-{{\log }_{3}}\frac{7}{9}=$

$\displaystyle {{\log }_{3}}\frac{7}{{\frac{7}{9}}}={{\log }_{3}}7\cdot \frac{9}{7}=$

$\displaystyle {{\log }_{3}}9=x$

Now the logarithm form we write it on an exponential form to find the answer

$\displaystyle {{3}^{x}}=9$

$\displaystyle {{3}^{x}}={{3}^{2}}$

$\displaystyle x=2$

So $\displaystyle {{\log }_{3}}7-{{\log }_{3}}\frac{7}{9}=2$

b) $\displaystyle {{\log }_{5}}100-{{\log }_{5}}4$

Finding each of them separately is difficult

$\displaystyle {{\log }_{5}}100={{x}_{1}}$

And $\displaystyle {{\log }_{5}}4={{x}_{2}}$

Based on the rule we know that the difference of the logarithms is equal with the quotient of logarithms

$\displaystyle {{\log }_{5}}100-{{\log }_{5}}4=$

$\displaystyle {{\log }_{5}}\frac{{100}}{4}=$

$\displaystyle {{\log }_{5}}25=x$

Now the logarithm form we write it on an exponential form to find the answer

$\displaystyle {{5}^{x}}=25$

$\displaystyle {{5}^{x}}={{5}^{2}}$

$\displaystyle x=2$

So $\displaystyle {{\log }_{5}}100-{{\log }_{5}}4=2$

Logarithm of a Power

The other property of exponents is power of a power and the similarity with logarithm property is that the power n becomes a factor.

$\displaystyle {{\log }_{a}}{{M}^{n}}=n{{\log }_{a}}M$

Example 5: Use the power property to simplify the logarithm.

a) $\displaystyle {{\log }_{2}}{{(32)}^{{12}}}$

Using the power property

$\displaystyle {{\log }_{2}}{{(32)}^{{12}}}=$

$\displaystyle 12{{\log }_{2}}(32)=$

Evaluating $\displaystyle {{\log }_{2}}(32)=x$

$\displaystyle {{2}^{x}}=32$

$\displaystyle {{2}^{x}}={{2}^{5}}$

$\displaystyle x=5$

So $\displaystyle {{\log }_{2}}{{(32)}^{{12}}}=$

$\displaystyle 12{{\log }_{2}}(32)=$

$\displaystyle 12\cdot 5=60$

b) $\displaystyle {{\log }_{9}}{{(81)}^{x}}$

Using the power property

$\displaystyle {{\log }_{9}}{{(81)}^{x}}=$

$\displaystyle x{{\log }_{9}}81=$

Evaluating $\displaystyle {{\log }_{9}}81=y$

$\displaystyle {{9}^{y}}=81$

$\displaystyle {{9}^{y}}={{9}^{2}}$

$\displaystyle y=2$

So $\displaystyle {{\log }_{9}}{{(81)}^{x}}=$

$\displaystyle x{{\log }_{9}}81=$

$\displaystyle x\cdot 2=2x$

Change of Base Property
The change of base property of logarithm says that we can rewrite a given logarithm as the ratio of two logarithms with any new base.
$\displaystyle {{\log }_{a}}x=\frac{{{{{\log }}_{b}}x}}{{{{{\log }}_{b}}a}}$

Example 6: Evaluate.

a) $\displaystyle {{\log }_{2}}7$

This type of logarithm is difficult to evaluate even with calculator, that’s why we use the change of base property to turn it with base 10.

$\displaystyle {{\log }_{2}}7=\frac{{{{{\log }}_{{10}}}7}}{{{{{\log }}_{{10}}}2}}=$

We evaluate each logarithm with calculator

$\displaystyle \frac{{\log 7}}{{\log 2}}\approx \frac{{0,84}}{{0,30}}\approx 2,8$

b) $\displaystyle {{\log }_{5}}8$

$\displaystyle {{\log }_{5}}8=\frac{{\log 8}}{{\log 5}}\approx \frac{{0,90}}{{0,70}}\approx 1,28$

Tip!

1. If a log has no base written, you should generally assume that the base is 10.

2. The logarithm to the base e ($\displaystyle {{\log }_{e}}x$) is written as $\displaystyle \ln x$.

Example 7: Simplify by using the rules of exponents and logarithms.

a) $\displaystyle \log \frac{1}{2}+\log \frac{2}{5}+\log \frac{5}{7}+\log \frac{7}{9}+\log \frac{9}{{10}}$

$\displaystyle \log \frac{1}{2}+\log \frac{2}{5}+\log \frac{5}{7}+\log \frac{7}{9}+\log \frac{9}{{10}}=$

Using the logarithm of a product rule

$\displaystyle \log (\frac{1}{2}\cdot \frac{2}{5}\cdot \frac{5}{7}\cdot \frac{7}{9}\cdot \frac{9}{{10}})=$

$\displaystyle \log (\frac{1}{{\cancel{2}}}\cdot \frac{{\cancel{2}}}{{\cancel{5}}}\cdot \frac{{\cancel{5}}}{{\cancel{7}}}\cdot \frac{{\cancel{7}}}{{\cancel{9}}}\cdot \frac{{\cancel{9}}}{{10}})=$

$\displaystyle \log (\frac{1}{{10}})=x$

$\displaystyle {{10}^{x}}=\frac{1}{{10}}$

$\displaystyle {{10}^{x}}={{10}^{{-1}}}$

$\displaystyle x=-1$

So $\displaystyle \log \frac{1}{2}+\log \frac{2}{5}+\log \frac{5}{7}+\log \frac{7}{9}+\log \frac{9}{{10}}=-1$

b) $\displaystyle {{e}^{{\ln 10-\ln 2}}}$

$\displaystyle {{e}^{{\ln 10-\ln 2}}}=$

Using Logarithm of a quotient rule

$\displaystyle {{e}^{{\ln (\frac{{10}}{2})}}}={{e}^{{\ln 5}}}=5$

c) $\displaystyle {{\log }_{2}}\frac{{8\sqrt{2}}}{{\sqrt[3]{2}}}$

Using the rules of exponents we rewrite our expression

$\displaystyle {{\log }_{2}}\frac{{8\sqrt{2}}}{{\sqrt[3]{2}}}={{\log }_{2}}\frac{{{{2}^{3}}\cdot {{2}^{{\frac{1}{2}}}}}}{{{{2}^{{\frac{1}{3}}}}}}=$

$\displaystyle {{\log }_{2}}{{2}^{{(3+\frac{1}{2}-\frac{1}{3})}}}={{\log }_{2}}{{2}^{{(\frac{{18+3-2}}{6})}}}=$

$\displaystyle {{\log }_{2}}{{2}^{{\frac{{19}}{6}}}}=\frac{{19}}{6}$

d) $\displaystyle {{\log }_{2}}\sqrt[5]{{8\sqrt[3]{2}}}$

Using the rules of exponents we rewrite our expression

$\displaystyle {{\log }_{2}}\sqrt[5]{{8\sqrt[3]{2}}}={{\log }_{2}}\sqrt[5]{{8\cdot {{2}^{{\frac{1}{3}}}}}}=$

$\displaystyle {{\log }_{2}}\sqrt[5]{{{{2}^{3}}\cdot {{2}^{{\frac{1}{3}}}}}}={{\log }_{2}}\sqrt[5]{{{{2}^{{(3+\frac{1}{3})}}}}}=$

$\displaystyle {{\log }_{2}}\sqrt[5]{{{{2}^{{(\frac{{9+1}}{3})}}}}}={{\log }_{2}}\sqrt[5]{{{{2}^{{(\frac{{10}}{3})}}}}}=$

$\displaystyle {{\log }_{2}}{{\left( {{{2}^{{(\frac{{10}}{3})}}}} \right)}^{{\frac{1}{5}}}}={{\log }_{2}}{{2}^{{(\frac{{10}}{3}\cdot \frac{1}{5})}}}=$

$\displaystyle {{\log }_{2}}{{2}^{{(\frac{{10}}{{15}})}}}={{\log }_{2}}{{2}^{{(\frac{2}{3})}}}=x$

$\displaystyle {{2}^{x}}={{2}^{{\frac{2}{3}}}}$ then $\displaystyle x=\frac{2}{3}$

So $\displaystyle {{\log }_{2}}\sqrt[5]{{8\sqrt[3]{2}}}=\frac{2}{3}$

e) $\displaystyle \frac{3}{{{{{\log }}_{2}}(2a)}}+\frac{3}{{{{{\log }}_{a}}(2a)}}$  where $\displaystyle a>1$

$\displaystyle \frac{3}{{{{{\log }}_{2}}(2a)}}+\frac{3}{{{{{\log }}_{a}}(2a)}}=$

Using the logarithm of a product rule

$\displaystyle \frac{3}{{{{{\log }}_{2}}2+{{{\log }}_{2}}a}}+\frac{3}{{{{{\log }}_{a}}2+{{{\log }}_{a}}a}}=$

We evaluate

$\displaystyle {{\log }_{2}}2=x$

$\displaystyle {{2}^{x}}={{2}^{1}}$

$\displaystyle x=1$

So $\displaystyle {{\log }_{2}}2=1$

In the same way $\displaystyle {{\log }_{a}}a=1$

Substituting we obtain

$\displaystyle \frac{3}{{1+{{{\log }}_{2}}a}}+\frac{3}{{1+{{{\log }}_{a}}2}}=$

Using the change of base rule

$\displaystyle {{\log }_{a}}2=\frac{{{{{\log }}_{2}}2}}{{{{{\log }}_{2}}a}}=\frac{1}{{{{{\log }}_{2}}a}}$

We get

$\displaystyle \frac{3}{{1+{{{\log }}_{2}}a}}+\frac{3}{{1+\frac{1}{{{{{\log }}_{2}}a}}}}=$

$\displaystyle \frac{3}{{1+{{{\log }}_{2}}a}}+\frac{3}{{\frac{{{{{\log }}_{2}}a+1}}{{{{{\log }}_{2}}a}}}}=$

$\displaystyle \frac{3}{{1+{{{\log }}_{2}}a}}+\frac{{3{{{\log }}_{2}}a}}{{1+{{{\log }}_{2}}a}}=$

$\displaystyle \frac{{3(1+{{{\log }}_{2}}a)}}{{1+{{{\log }}_{2}}a}}=$

$\displaystyle \frac{{3\cancel{{(1+{{{\log }}_{2}}a)}}}}{{\cancel{{1+{{{\log }}_{2}}a}}}}=3$

So $\displaystyle \frac{3}{{{{{\log }}_{2}}(2a)}}+\frac{3}{{{{{\log }}_{a}}(2a)}}=3$

f) $\displaystyle {{3}^{{\frac{2}{{{{{\log }}_{5}}81}}}}}$

Using the change of base rule

$\displaystyle {{\log }_{5}}3=\frac{{{{{\log }}_{3}}3}}{{{{{\log }}_{3}}5}}=\frac{1}{{{{{\log }}_{3}}5}}$

$\displaystyle {{3}^{{\frac{2}{{{{{\log }}_{5}}81}}}}}={{3}^{{\frac{2}{{{{{\log }}_{5}}{{3}^{4}}}}}}}={{3}^{{\frac{2}{{4{{{\log }}_{5}}3}}}}}$

Substituting we obtain

$\displaystyle {{3}^{{\frac{2}{{\frac{4}{{{{{\log }}_{3}}5}}}}}}}={{3}^{{\frac{{2{{{\log }}_{3}}5}}{4}}}}={{3}^{{\frac{{{{{\log }}_{3}}5}}{2}}}}$

Using logarithm of a power rule

$\displaystyle {{3}^{{{{{\log }}_{3}}{{5}^{{\frac{1}{2}}}}}}}={{5}^{{\frac{1}{2}}}}=\sqrt{5}$

So $\displaystyle {{3}^{{\frac{2}{{{{{\log }}_{5}}81}}}}}=\sqrt{5}$

g) $\displaystyle {{\log }_{{\frac{1}{9}}}}27-{{\log }_{{81}}}\frac{1}{3}$

$\displaystyle {{\log }_{{\frac{1}{9}}}}27-{{\log }_{{81}}}\frac{1}{3}=$

We can’t use the quotient rule because the bases are different.

Let’s evaluate them separately

$\displaystyle {{\log }_{{\frac{1}{9}}}}27=x$

$\displaystyle {{\left( {\frac{1}{9}} \right)}^{x}}=27$

$\displaystyle {{\left( {\frac{1}{{{{3}^{2}}}}} \right)}^{x}}=27$

$\displaystyle {{\left( {{{3}^{{-2}}}} \right)}^{x}}={{3}^{3}}$

$\displaystyle {{3}^{{-2x}}}={{3}^{3}}$ then $\displaystyle -2x=3$

$\displaystyle x=-\frac{3}{2}$

$\displaystyle {{\log }_{{81}}}\frac{1}{3}=y$ then $\displaystyle {{81}^{y}}=\frac{1}{3}$

$\displaystyle {{3}^{{4y}}}={{3}^{{-1}}}$ then $\displaystyle 4y=-1$

$\displaystyle y=-\frac{1}{4}$

Now we substitute and evaluate

$\displaystyle x-y=-\frac{3}{2}-\left( {-\frac{1}{4}} \right)=$

$\displaystyle -\frac{6}{4}+\frac{1}{4}=-\frac{5}{4}$

So $\displaystyle {{\log }_{{\frac{1}{9}}}}27-{{\log }_{{81}}}\frac{1}{3}=-\frac{5}{4}$

Remember!

$\displaystyle {{\log }_{b}}1=0$ because $\displaystyle {{b}^{0}}=1$

$\displaystyle {{\log }_{b}}b=1$ because $\displaystyle {{b}^{1}}=b$

$\displaystyle {{b}^{{{{{\log }}_{b}}x}}}=x$

$\displaystyle {{\log }_{b}}{{b}^{x}}=x$

Example 8: Find x.

a) $\displaystyle {{\log }_{2}}({{\log }_{3}}x)=5$

$\displaystyle {{\log }_{3}}x={{2}^{5}}$

$\displaystyle {{\log }_{3}}x=32$

$\displaystyle x={{3}^{{32}}}$

b) $\displaystyle {{25}^{{{{{\log }}_{5}}x}}}=x+56$

We should note that x>0

$\displaystyle {{5}^{{2{{{\log }}_{5}}x}}}=x+56$

$\displaystyle {{5}^{{{{{\log }}_{5}}{{x}^{2}}}}}=x+56$

$\displaystyle {{x}^{2}}=x+56$

$\displaystyle {{x}^{2}}-x-56=0$

$\displaystyle {{x}_{1}}=-7$  and $\displaystyle {{x}_{2}}=8$

c) $\displaystyle {{4}^{{\log x}}}\cdot {{2}^{{\log x}}}=64$

$\displaystyle {{2}^{{2\log x}}}\cdot {{2}^{{\log x}}}=64$

$\displaystyle {{2}^{{2\log x+}}}^{{\log x}}=64$

$\displaystyle {{2}^{{3\log x}}}={{2}^{6}}$

$\displaystyle 3\log x=6$

$\displaystyle \log x=2$

$\displaystyle x={{10}^{2}}$

$\displaystyle x=100$

d) $\displaystyle {{\log }_{{\frac{1}{3}}}}x=\frac{2}{{{{{\log }}_{3}}5}}$

$\displaystyle x={{\left( {\frac{1}{5}} \right)}^{{\frac{2}{{{{{\log }}_{3}}5}}}}}=$

$\displaystyle {{({{5}^{{-1}}})}^{{^{{\frac{2}{{{{{\log }}_{3}}5}}}}}}}=$

$\displaystyle {{(5)}^{{-\frac{2}{{{{{\log }}_{3}}5}}}}}=$

Changing the base rule

$\displaystyle {{\log }_{3}}5=\frac{{{{{\log }}_{5}}5}}{{{{{\log }}_{5}}3}}=\frac{1}{{{{{\log }}_{5}}3}}$

Substituting we obtain

$\displaystyle {{(5)}^{{\frac{{-2}}{{\frac{1}{{{{{\log }}_{5}}3}}}}}}}=$

$\displaystyle {{5}^{{-2{{{\log }}_{5}}3}}}=$

$\displaystyle {{5}^{{{{{\log }}_{5}}{{{(3)}}^{{-2}}}}}}=$

$\displaystyle {{3}^{{-2}}}=\frac{1}{9}$

So $\displaystyle x=\frac{1}{9}$

e) $\displaystyle {{9}^{{\ln x}}}-4\cdot {{3}^{{1+\ln x}}}+27=0$

We should note that x>0

$\displaystyle {{3}^{{2\ln x}}}-4\cdot {{3}^{1}}\cdot {{3}^{{\ln x}}}+27=0$

$\displaystyle {{3}^{{2\ln x}}}-12\cdot {{3}^{{\ln x}}}+27=0$

We substitute $\displaystyle {{3}^{{\ln x}}}=t$ and obtain a quadratic equation of the form $\displaystyle {{t}^{2}}-12t+27=0$

We find $\displaystyle {{t}_{1}}=3$ and $\displaystyle {{t}_{2}}=9$

Now we find the x-s

$\displaystyle {{3}^{{\ln {{x}_{1}}}}}=3$

$\displaystyle \ln {{x}_{1}}=1$

$\displaystyle {{x}_{1}}=e$ and $\displaystyle {{3}^{{\ln {{x}_{2}}}}}=9$

$\displaystyle {{3}^{{\ln {{x}_{2}}}}}={{3}^{2}}$

$\displaystyle \ln {{x}_{2}}=2$

$\displaystyle {{x}_{2}}={{e}^{2}}$

f) $\displaystyle {{({{\log }_{5}}20)}^{2}}={{(\log {{5}^{4}})}^{2}}+{{\log }_{5}}x$

$\displaystyle {{\log }_{5}}x={{({{\log }_{5}}20)}^{2}}-{{({{\log }_{5}}4)}^{2}}=$

$\displaystyle ({{\log }_{5}}20-{{\log }_{5}}4)({{\log }_{5}}20+{{\log }_{5}}4)$

Using the product and quotient rule

$\displaystyle ({{\log }_{5}}\frac{{20}}{4})({{\log }_{5}}20\cdot 4)=$

$\displaystyle ({{\log }_{5}}5)({{\log }_{5}}80)={{\log }_{5}}80$

$\displaystyle x=80$