Question

2log(2x)base at 8+log(x-1)2 base at 8=4/3

Answer 1

Answer:

x is 2.

Step by step Explanation:

$2 log_{8}(2x) + log_{8}(( {x - 1})^{2} ) = \frac{4}{3} \\ = > 2 log_{8}(2x) + 2 log_{8}(( {x - 1})) = \frac{4}{3} \\ = > 2( log_{8}(2x) + log_{8}(( {x - 1}) ) )= \frac{4}{3} \\ = > ( log_{8}(2x) + log_{8}(( {x - 1}) ) )= \frac{2}{3} \\ = > log_{8}(2x(x - 1)) = \frac{2}{3} \\ = > (2x(x - 1)) = {8}^{ \frac{2}{3} } \\ = > 2 ({x}^{2} - x) = { {2}^{3} }^{ \frac{2}{3} } = {2}^{3 \times \frac{2}{3} } = {2}^{2} = 4 \\ = > 2 ({x}^{2} - x) = 4 \\ = > {x}^{2} - x = 2 \\ = > {x}^{2} - x - 2 = 0 \\ = > {x}^{2} - 2x + x - 2 = 0 \\ = > x( x - 2) + 1(x - 2) = 0 \\ = > (x - 2)(x + 1) = 0 \\ = > x = 2 \: or \: - 1 \\ (as \: log \: of \: negative \: value \: is \: invalid \\ \: so \: x \: can \: not \: be \: - 1) \\ = so \: x \: is \: 2(ans.)$

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