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If $ \dfrac{{\log 64}}{{\log 8}} = \log x, $ then value of x is
A. $ 100 $
B. $ 112 $
C. $ 165 $
D. $ 95 $

Last updated date: 13th Jun 2024
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Hint: Logarithms are ways to figure out which exponents we need to multiply into the specific number. Here, using the property of logarithm the change of base, According to the power rule, the log of a power is equal to the power times the log of the base.
 $ \log {}_b{a^N} = N{\log _b}a $

Complete step-by-step answer:
Take the given expression –
 $ \dfrac{{\log 64}}{{\log 8}} = \log x $
Now, the number can be written as the square of number. $ 64 = {8^2} $
 $ \Rightarrow \dfrac{{\log {8^2}}}{{\log 8}} = \log x $
By using the power rule - the log of a power is equal to the power times the log of the base.
 $ \log {}_b{a^N} = N{\log _b}a $
 $ \Rightarrow \dfrac{{2\log 8}}{{\log 8}} = \log x $
Common multiples from the numerator and denominator cancel each other.
 $ \Rightarrow 2 = \log x $
Let us assume the base $ 10 $ of the logarithm.
Simplify the above equation –
 $ \Rightarrow x = {10^2} $
Simplify the above equation by placing the square of the number. As we know that $ {10^2} = 100 $
 $ \Rightarrow x = 100 $
Hence from the given multiple choices – the option A is the correct answer.
So, the correct answer is “Option A”.

Note: In other words, the logarithm is the power to which the number must be raised in order to get some other. Always remember the standard properties of the logarithm.... Product rule, quotient rule and the power rule. The basic logarithm properties are most important and the solution solely depends on it, so remember and understand its application properly. Be good in multiples and know the concepts of square and square root and apply accordingly.
Also refer to the below properties and rules of the logarithm.
Product rule: $ {\log _a}xy = {\log _a}x + {\log _a}y $
Quotient rule: $ {\log _a}\dfrac{x}{y} = {\log _a}x - {\log _a}y $
Power rule: $ {\log _a}{x^n} = n{\log _a}x $
 Base rule: $ {\log _a}a = 1 $
Change of base rule: $ {\log _a}M = \dfrac{{\log M}}{{\log N}} $