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Logarithms Practice Test 1

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Question : 14 of 40

Marks: +1, -0

If

2 \log_{16} (x^2 + x) - \log_{4} (x + 1) = 2

, then x = ?

– 1
16
2
None of these

Solution: 👈: Video Solution

Concept:

Use change of base and logarithm laws to reduce the equation to a simple form.

Explanation:

Given:

2\log_{16}(x^2+x)-\log_4(x+1)=2

.

Since

16=4^2

, we have

\log_{16}A=\frac{\log_4 A}{2}

.

So

2\log_{16}(x^2+x)=\log_4(x^2+x)

.

The equation becomes

\log_4(x^2+x)-\log_4(x+1)=2

.

Using

\log a-\log b=\log\left(\frac{a}{b}\right)

:

\log_4\left(\frac{x^2+x}{x+1}\right)=2

.

Now

\frac{x^2+x}{x+1}=\frac{x(x+1)}{x+1}=x

for

x\neq -1

.

Thus

\log_4 x=2

.

So

x=4^2=16

.

Check domain:

x>0

, so

x=16

is valid.

Answer:

x=16

, so option B.

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