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Question: Find x, if \({4^{{{\log }_9}3}} + {9^{{{\log }_2}4}} = {10^{{{\log }_x}83}}\)...

Find x, if 4log⁡93+9log⁡24=10log⁡x83{4^{{{\log }_9}3}} + {9^{{{\log }_2}4}} = {10^{{{\log }_x}83}}

Explanation

Solution

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⁡baN=Nlog⁡ba{\log _b}{a^N} = N{\log _b}a and log⁡bNa=1Nlog⁡ba{\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, 4log⁡93+9log⁡24=10log⁡x83{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}aandand{\log _{{b^N}}}a = \dfrac{1}{N}{\log _b}aNowtakingexponentof 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,tosolvetheaboveexpression,wehaveused Here, to solve the above expression, we have used{\log _{{b^N}}}a = \dfrac{1}{N}{\log _b}a.Similarly,takingexponentof. 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 = 2Here,tosolvetheaboveexpression,wehaveused 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

{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.