Question

If , $\mathrm x-\log48+3\log2=\frac13\log125-\log3\;\mathrm{and}\;\mathrm x=\log_{\mathrm m}20$, then the value of m will be-
(A). 6
(B). 5
(C). 10
(D). 20

Answer 1

Hint: In this question, the properties of logarithmic functions will be used. Some of the properties are-
$\mathrm{bloga}=\mathrm{loga}^{\mathrm b}.....\left(1\right)\\\\\log_{\mathrm m}\mathrm a=\dfrac{\mathrm{loga}}{\mathrm{logm}}.....\left(2\right)\\\\\mathrm{logab}=\mathrm{loga}+\mathrm{logb}.....\left(3\right)\\\\\log_{\mathrm a}\mathrm b=\mathrm c\;\mathrm{is}\;\mathrm{the}\;\mathrm{same}\;\mathrm{as}\;\mathrm a^{\mathrm c}=\mathrm b....\left(4\right)$
Also, the logarithm base is considered to be 10.

Complete step-by-step answer:
Using the above mentioned properties, we can easily solve the logarithmic equation-
$\mathrm x-\log48+3\log2=\dfrac13\log125-\log3\\\\\mathrm x-\log\left(2^4\times3\right)+3\log2=\log125^\frac13-\log3\\\\\mathrm{Using}\;\mathrm{properties}\;\left(1\right)\;\mathrm{and}\left(3\right),\\\\\mathrm x-4\log2-\log3+3\log2=\log5-\log3\\\\\mathrm x=\log5+\log2=\log10\\\\\mathrm{But}\;\mathrm x=\log_{\mathrm m}20\;\mathrm{and}\;\log10=1,\\\\\log_{\mathrm m}20=1\\\\\mathrm{Using}\;\mathrm{property}\;\left(4\right),\\\\\mathrm m^1=20\\\\\mathrm m=20$
Hence, the correct answer is D. 20

Note: In such problems first simplify the larger logarithms into log of prime numbers. This helps in solving the value of variables easily. The base of the log in the question is 10, and it can be manipulated according to the requirement of the question.