Question

"1 + log_{2}(x - 2) = log_{2}x" alt="1 + log_{2}(x - 2) = log_{2}x" align="absmiddle" class="latex-formula">

Answer 1

Move all the logarithms on the left hand side, and all the constants on the other:

$\log_2(x-2) - \log_2(x) = -1$

Use the rule of logarithms

$\log_a(b) - \log_a(c) = \log_a\left(\dfrac{b}{c}\right)$

To rewrite the equation as

$\log_2\left(\dfrac{x-2}{x}\right) = -1$

Evaluate 2 to the power of each side:

$\dfrac{x-2}{x} = 2^{-1} = \dfrac{1}{2}$

Multiply both sides by 2x:

$2(x-2) = x \iff 2x-4 = x \iff x = 4$

Answer 2

1 + log₂(x - 2) = log₂(x)

1 = log₂(x) - log₂(x - 2)

1 = log₂$\frac{x}{x - 2}$

2¹ = $\frac{x}{x - 2}$

2(x - 2) = x

2x - 4 = x

    -4 = -x

     4 = x

Answer: 4