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Question

Question: How do you evaluate \( {\log _{15}}\left( {15} \right)? \)...

How do you evaluate log15(15)?{\log _{15}}\left( {15} \right)?

Explanation

Solution

Hint : As we know that the logarithm is the inverse function to exponentiation. That means the logarithm of a given number xx is the exponent to which another fixed number, the base b, must be raised , to produce that number xx . As per the definition of a logarithm logab=c{\log _a}b = c which gives that ac=b{a^c} = b . Here in the above expression the base is 1515 . And we also have to assume that if no base bb is written then the base is always 10. This is an example of base ten logarithm because 1010 is the number that is raised to a power.

Complete step-by-step answer :
As per the given question we have log15(15){\log _{15}}\left( {15} \right) . We know that if
logab=x{\log _a}b = x , then ax=b{a^x} = b .
Let us take log15(15)=x{\log _{15}}\left( {15} \right) = x .
As we know that the inverse function of log15(z)=15z{\log _{15}}(z) = {15^z} and also it means that log15(15z)=z{\log _{15}}({15^z}) = z and also if there is 15log15(z){15^{{{\log }_{15}}(z)}} is equal to zz .
So to here find the value of xx , we need to get rid of the logarithm term, we will apply this rule, here z=15z = 15 .
Therefore 15log15(15)=15x{15^{{{\log }_{15}}(15)}} = {15^x} , By applying the above rule we can write
15=15x\Rightarrow 15 = {15^x} .
It gives us 151=15xx=1{15^1} = {15^x} \Rightarrow x = 1 .
Hence the value of log15(15)=1{\log _{15}}\left( {15} \right) = 1 .
So, the correct answer is “1”.

Note : We should always be careful while solving logarithm formulas and before solving this kind of problems we should know all the rules of logarithm and exponentiation. We have to keep in mind that when a logarithm is written without any base, like this: log100\log 100 then this usually means that the base is already there which is 1010 . It is called a common logarithm or decadic logarithm, is the logarithm to the base 1010 . One way we can approach log problems is to keep in mind that ab=c{a^b} = c and logac=b{\log _a}c = b .