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# Solve the following equation having given log2, log3 and log7, ${21^x} = {2^{2x + 1}}{.5^x}$

Last updated date: 16th Sep 2024
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Answer
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Hint: To solve this question, we will use some of the logarithmic properties, i.e. $\log {a^b} = b\log a,\log \left( {ab} \right) = \log a + \log b$ etc. using these properties, we can solve this.

Complete step-by-step answer:
We have,
${21^x} = {2^{2x + 1}}{.5^x}$ …… (i)
Taking log both sides,
$\Rightarrow \log \left( {{{21}^x}} \right) = \log \left( {{2^{2x + 1}}{{.5}^x}} \right)$
We know that,
$\Rightarrow \log {a^b} = b\log a$ and $\log \left( {ab} \right) = \log a + \log b$
So,
$\Rightarrow x\log \left( {21} \right) = \log \left( {{2^{2x + 1}}} \right) + \log \left( {{5^x}} \right)$
$\Rightarrow x\log \left( {21} \right) = \left( {2x + 1} \right)\log \left( 2 \right) + x\log \left( 5 \right)$
We can also write this as,
$\Rightarrow x\log \left( {7.3} \right) = \left( {2x + 1} \right)\log \left( 2 \right) + x\log \left( 5 \right)$
It will become,
$\Rightarrow x\log \left( 7 \right) + x\log \left( 3 \right) = \left( {2x} \right)\log \left( 2 \right) + \log \left( 2 \right) + x\log \left( 5 \right)$
Separating the coefficients of x,
$\Rightarrow x\left[ {\log \left( 7 \right) + \log \left( 3 \right) - 2\log \left( 2 \right) - \log \left( 5 \right)} \right] = \log \left( 2 \right)$
$\Rightarrow x = \dfrac{{\log \left( 2 \right)}}{{\left[ {\log \left( 7 \right) + \log \left( 3 \right) - 2\log \left( 2 \right) - \log \left( 5 \right)} \right]}}$ …….. (ii)
As we know that, log10 =1
We can write log10 as:
$\Rightarrow \log \left( {5.2} \right) = 1$
$\Rightarrow \log \left( 5 \right) + \log \left( 2 \right) = 1$
$\Rightarrow \log \left( 5 \right) = 1 - \log \left( 2 \right)$
Putting this value in equation (ii),
$\Rightarrow x = \dfrac{{\log \left( 2 \right)}}{{\left[ {\log \left( 7 \right) + \log \left( 3 \right) - 2\log \left( 2 \right) - 1 + \log \left( 2 \right)} \right]}}$
$\Rightarrow x = \dfrac{{\log \left( 2 \right)}}{{\left[ {\log \left( 7 \right) + \log \left( 3 \right) - \log \left( 2 \right) - 1} \right]}}$
Here, we can see that the value of x is in the form having the given log2, log3 and log7.
Hence, it is the correct answer.

Note: Y times the logarithm of x is the logarithm of the exponent x, elevated to the power of y. The power rule can be used by multiplication operation for quick exponent calculation. Where appropriate, use the correct formula or logarithm and try to solve the problem.