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Logarithms Practice Test 2

Question : 21 of 39

Marks: +1, -0

\frac{1}{\log_{a} x} + \frac{1}{\log_{c} x} = \frac{2}{\log_{b} x}

, then a, b, c are in:

Solution:

Concept:

Use the reciprocal-log identity

\frac{1}{\log_a x}=\log_x a

to simplify the equation.

Explanation:

Given:

\frac{1}{\log_a x}+\frac{1}{\log_c x}=\frac{2}{\log_b x}

Apply the identity to each term:

\log_x a+\log_x c=2\log_x b

Use the product and power rules:

\log_x(ac)=\log_x(b^2)

Since the logarithms have the same base, their arguments must be equal:

ac=b^2

b

is the geometric mean of

a

and

c

Hence,

a

b

, and

c

are in G.P.

Answer:

Option B: G.P.

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