What is the value of the logarithmic expression \frac{1}{\log_3 12} + \frac{1}{\log_4 12}? — JAMB UTME

What is the value of the logarithmic expression $\frac{1}{\log_3 12} + \frac{1}{\log_4 12}$ ?

1

Using the change of base relationship

\frac{1}{\log_a b} = \log_b a

, the expression simplifies to

\log_{12} 3 + \log_{12} 4

. By the product rule of logarithms, this equals

\log_{12}(3 \times 4) = \log_{12} 12 = 1

Apply the change of base rule

\frac{1}{\log_a b} = \log_b a

to each term.

\frac{1}{\log_3 12} = \log_{12} 3 \quad \text{and} \quad \frac{1}{\log_4 12} = \log_{12} 4

Expressing both terms with a common base of

12

enables the use of logarithmic laws.

Apply the product law of logarithms

\log_b M + \log_b N = \log_b (M \times N)

\log_{12} 3 + \log_{12} 4 = \log_{12} (3 \times 4) = \log_{12} 12

The sum of logarithms with identical bases equals the logarithm of the product of their arguments.

Simplify

\log_{12} 12

The logarithm of any base to itself is always

1

(

\log_a a = 1

Change of Base Property and Logarithm Addition Law

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