Question

How do you solve${23^x} = 6$?

Answer 1

Hint:
Whenever we have such types of problems where we are given any number with the power of any variable like here we have $x$ we just need to take the logarithm to both the sides and apply its property to solve accordingly in order to find the value of the unknown term.

Complete step by step solution:
Here we are given to solve ${23^x} = 6$ which means we need to find the value of $x$
Here we do not have any direct method to find its value as it is in the power of any number. So here we need to take the logarithm to both sides and it can be taken to any base as it will be the same on both the sides in LHS as well as in the RHS.
So we are given to solve the term:
${23^x} = 6$$ - - - (1)$
Now taking $\log $to both sides we will get:
$\log \left( {{{23}^x}} \right) = \log \left( 6 \right)$$ - - - - (2)$
Now here we could also have taken the logarithm to the base $e$ on both sides the answer would be the same.
Now we need to apply the property of the logarithm which says that:
$\log {m^n} = n\log m$
So we can also substitute in above $m = 23,n = x$
We can write:
$\log {23^x} = x\log 23$
Now substituting this value in equation (2) we will get:
$x\log \left( {23} \right) = \log \left( 6 \right)$
So we get $x = \dfrac{{\log 6}}{{\log 23}}$

We can also write it as ${\log _{23}}6$

Note:
Here in such problems the student must know the properties of the log properly because without that these types of problems cannot be solved. Hence the properties of logarithm are like:
$\log \left( {ab} \right) = \log \left( a \right) + \log \left( b \right)$
$\log \left( {\dfrac{a}{b}} \right) = \log \left( a \right) - \log \left( b \right)$
$\log {m^n} = n\log m$
Hence these need to be kept in mind to solve such problems.