Question

Find x, if ${4^{{{\log }_9}3}} + {9^{{{\log }_2}4}} = {10^{{{\log }_x}83}}$

Answer 1

Hint: To solve this question, i.e., to find x, we will start with solving exponents first from the L.H.S., by using basic log identities, ${\log _b}{a^N} = N{\log _b}a$ and ${\log _{{b^N}}}a = \dfrac{1}{N}{\log _b}a$. Then taking R.H.S. as it is, we will equate with the earlier solved L.H.S. expression, on solving further we will get the value of x.

Complete step-by-step answer:
We have been given an equation, ${4^{{{\log }_9}3}} + {9^{{{\log }_2}4}} = {10^{{{\log }_x}83}}$. We need to find x.
So, ${4^{{{\log }_9}3}} + {9^{{{\log }_2}4}} = {10^{{{\log }_x}83}}$\[ \ldots .eq.\left( 1 \right)\]
We know that, ${\log _b}{a^N} = N{\log _b}a$ and ${\log _{{b^N}}}a = \dfrac{1}{N}{\log _b}a$
Now taking exponent of ${4^{{{\log }_9}3}}$ from L.H.S. of \[eq.\left( 1 \right)\] to solve separately, we get
${\log _9}3 = {\log _{{3^2}}}3 = \dfrac{1}{2}{\log _3}3 = \dfrac{1}{2}$
Here, to solve the above expression, we have used ${\log _{{b^N}}}a = \dfrac{1}{N}{\log _b}a$.
Similarly, taking exponent of ${9^{{{\log }_2}4}}$ from L.H.S. of \[eq.\left( 1 \right)\] to solve separately, we get
${\log _2}4 = {\log _2}{2^2} = 2{\log _2}2 = 2$
Here, to solve the above expression, we have used ${\log _b}{a^N} = N{\log _b}a$.
Now, on putting both the above solved values in eq.(1), we get
\[\begin{gathered}
  {4^{\dfrac{1}{2}}} + {9^2} = {10^{{{\log }_x}83}} \\
  2 + 81 = {10^{{{\log }_x}83}} \\
\end{gathered} \]
\[83 = {10^{{{\log }_x}83}}\]
Now, taking ${\log _{10}}$ on both sides in the above equation, we get
\[{\log _{10}}83 = {\log _{10}}{10^{{{\log }_x}83}}\]
Now using, ${\log _b}{a^N} = N{\log _b}a$, we get
\[{\log _{10}}{10^{{{\log }_x}83}} = {\log _x}83{\log _{10}}{10^{}}\]
Then on putting in the equation, we get \[{\log _{10}}83 = {\log _x}83{\log _{10}}{10^{}}\]
We know that, \[{\log _{10}}10 = 1\]
Then, \[{\log _{10}}83 = {\log _x}83\]
On equating base value of log from both sides, we get
$\begin{gathered}
   \Rightarrow 10 = x \\
  x = 10 \\
\end{gathered} $
Thus, the value of x is \[10.\]

Note: Logarithm is actually the inverse function of exponentiation. Now let us understand with the help of an example.
Example of exponential function: To express “\[2\] raised to \[{3^{rd}}\] power equals to \[8\]” we write this as ${2^3} = 8$
Example of logarithm function: To express “\[2\]raised to which power equals to \[8\]?” we write this as \[lo{g_2}(8) = 3\], it reads as “log base two of eight is three”.
So, ${2^3} = 8$$ \Leftrightarrow $\[lo{g_2}(8) = 3\]
Thus, logarithm is the inverse function of exponentiation.