Question

If m = log20 and n = log25, then the value of x, so that $2\log \left( x-4 \right)=2m-n$ is equal to
(a). 6
(b). 8
(c). 10
(d). 12

Answer 1

Hint: Substitute the value of m and n in the given logarithmic expression , Now use property of log such as ${{\log }_{b}}\left( \dfrac{M}{N} \right)={{\log }_{b}}M-{{\log }_{b}}N$ and ${{\log }_{b}}\left( {{M}^{k}} \right)=k{{\log }_{b}}M$ to obtain a quadratic equation in x , solve the obtained equation to get the answer.

Complete step-by-step answer:

It is given that m = log20 and n = log25
Let us consider the logarithmic equation,
$2\log \left( x-4 \right)=2m-n$
Put the value of the m and n, we get
$2\log \left( x-4 \right)=2\log 20-\log 25$
By using rule, the logarithm of the exponential number
$\log {{\left( x-4 \right)}^{2}}=\log {{20}^{2}}-\log 25$
By using rule, the logarithm of the quotient
$\log {{\left( x-4 \right)}^{2}}=\log \left( \dfrac{{{20}^{2}}}{25} \right)$
Cancelling the logarithm, we get
${{\left( x-4 \right)}^{2}}=\left( \dfrac{{{20}^{2}}}{25} \right)$
${{\left( x-4 \right)}^{2}}=\left( \dfrac{20\times 20}{25} \right)=\left( \dfrac{4\times 5\times 4\times 5}{5\times 5} \right)=4\times 4$
${{\left( x-4 \right)}^{2}}={{4}^{2}}$
Taking the squaring root on both sides, we get
$\left( x-4 \right)=4$
$x=4+4=8$
Hence, the required value of x is 8.
Therefore, the correct option is (b).

Note: The possibility for the mistake is that you might get confused about the difference between the $\log (x)$ and $\ln (x)$. Where $\log (x)$ is the logarithm to the base 10 and $\ln (x)$ is the logarithm to the base e.