If a² + c² = b², then {1}{_{b-c} a} + {1}{_{b+c} a} = 2 is true for:

Correct Answer is : b > |c| , ≠ |c| + 1, a > 0, ≠ 1

b > |c| , ≠ |c| + 1, a > 0, ≠ 1

Correct Answer is : b > |c| , ≠ |c| + 1, a > 0, ≠ 1

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Logarithms Practice Test 2

Question : 39 of 39

Marks: +1, -0

a^2 + c^2 = b^2

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

is true for:

Solution:

loga x is defined in real domain

x > 0, a > 0, \neq 1

\therefore \frac{1}{\log_{b-c} a} + \frac{1}{\log_{b+c} a}

is defined for b – c > 0, b + c > 0, b – c ≠ 1, b + c ≠ 1

∴ b > |c| and b ≠ |c| + 1.

Also for a = 1, we get denominator = 0

So, a > 0, ≠ 1.

∴ If b > |c| , ≠ |c| + 1 and a > 0, ≠ 1 , then

\frac{1}{\log_{b-c} a} + \frac{1}{\log_{b+c} a}

= \log_a (b - c) + \log_a (b + c)

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