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Logarithms Practice Test 2

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Question : 12 of 39

Marks: +1, -0

If

\log_4{x^2(x-a)^2} - \log_2(x-1) = 1,

then the value of x will be:

1
2
3
4

Solution:

Concept:

The base-change rule

\log_{b^n} m = \dfrac{1}{n}\log_b m

and the quotient rule

\log_b A - \log_b B = \log_b\dfrac{A}{B}

simplify the equation to a single logarithm.

Explanation:

Given:

\log_4 x^2(x-1)^2 - \log_2(x-1) = 1

Step 1: Write the power as a square of a product.

\log_4 [x(x-1)]^2 - \log_2(x-1) = 1

Step 2: Use

\log_{2^2} M^2 = \dfrac{2}{2}\log_2 M = \log_2 M

.

\log_2 [x(x-1)] - \log_2(x-1) = 1

Step 3: Apply the quotient rule for logarithms.

\log_2 \dfrac{x(x-1)}{x-1} = 1

Step 4: Cancel the common factor

(x-1)

.

\log_2 x = 1

Step 5: Convert to exponential form.

x = 2^1 = 2

Check:

x = 2 > 1

, so both logarithms are defined.

Answer:

x = 2

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