Question

Evaluate the given logarithmic expression: ${\log _{25}}5$

Answer 1

We are given a logarithmic function which has a base of 25 and we have to evaluate its value. For this we use the conversion of logarithms into exponents because both are inverse entities of each other and then by comparison of indices we will obtain our result.

Complete solution step by step:
Firstly we write the given logarithmic expression
${\log _{25}}5$

Now we look what exponents actually mean

Exponent of a number means how many times the number is multiplied by itself i.e. ${p^q} = p \times p \times p \times p........(q\,{\text{times}}) = r$

It says $p$multiplied by itself $q$times equals to $r$

And logarithms are just opposite to it where the following function
${\log _a}c = b\,{\text{ - - - - - - - - equation(1)}}$

Means that - When $a$ is multiplied by itself $b$number of times $c$ is obtained.

So we assume a value of our expression like this
${\log _{25}}5 = x$

Now this is an equation of the same type as equation (1) so by translating the definition of (1) here, we get
25 multiplied by itself $x$ times to obtain 5
$\Rightarrow {(25)^x} = 5\,{\text{ - - - - - - equation(2)}}$

Now using the following formula we convert this equation (2) like this
$[{\log _{25}}5 = x] \Leftrightarrow [{25^x} = 5]$

This means we have to select a power of 25 to get 5 so we convert 25 into 5 using the exponent i.e.
$25 = {5^2}$

Putting this value in equation (2) we get
${\left( {25} \right)^x} = {\left( {{5^2}} \right)^x} = {\left( 5 \right)^{2x}} = 5$

Now by using the following property of indices we have

${p^q} = {p^r} \\\ \Rightarrow q = r \\\$

Applying this in our problem
${\left( 5 \right)^{2x}} = \left( 5 \right) \\\ \Rightarrow 2x = 1 \\\ \Rightarrow x = \dfrac{1}{2} \\\$

This means our evaluated answer for the given expression is
${\log _{25}}5 = \dfrac{1}{2}$

Additional information: Exponential function is Inverse function of a logarithmic function. This means one can be undone or removed by operating the other function on it and vice versa. This would give you a better understanding of it –
${\log _a}c = b \\\ {a^b} = {a^{{{\log }_a}\;c}} = c \\\$

Doing the opposite will give us –
${a^b} = c \\\ {\log _a}c = {\log _a}({a^b}) = b \\\$

This helps us to understand the reason why they are inverse functions with each other.

Note: To evaluate a value of logarithm, we try to reach from bottom to top of a logarithmic function by using some number and operation. Here we had 25 as our base and we needed to get 5 so we took the square root of 25 i.e. power of $\dfrac{1}{2}$ to get our result and hence this index of $\dfrac{1}{2}$ becomes our answer.