Logarithm
Logarithm Properties
There is a close relationship between logarithms and exponents. 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 exponents then also their properties are similar.
Logarithm of a Product
With exponents 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$
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}}$
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$
Check out how to solve quadratic equations
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$
