Question

Find the value of ‘x’ if 3log 2 + $\dfrac{1}{3}$log27 – log4 = log x.

Hint: To solve the logarithmic equation given in the question we use the formulae for logarithmic functions and simplify for the answer.

Given data, 3log 2 + $\dfrac{1}{3}$log27 – log4 = log x
⟹3log 2 +$\dfrac{1}{3}$ log ${3^3}$ – log ${2^2}$ = log x (27 = ${3^3}$ and 4 =${2^2}$)
⟹3 log 2 + ($\dfrac{1}{3}$) × 3 log 3 – 2log2 = log x -- (log ${{\text{x}}^{\text{y}}}$ = y log x)
⟹3 log 2 + log 3 – 2log 2 = log x -- (log (x y) = log x + log y)
⟹log 2 + log 3 = log x
⟹log (2 x 3) = log x
⟹log 6 = log x
⟹ x = 6
Hence x = 6

Note: The key in solving such types of problems is to modify the equation using the logarithmic formulae.
Logarithm is the inverse function to exponentiation.
It is defined as a quantity representing the power to which a fixed number (the base) must be raised to produce a given number.