What is the solution of the equation $\log_2\dfrac{\sqrt{3x+4}}{x-2}= 1$ ?

$0$

$\dfrac{3}{4}$

$2$

$4$

$7$

Approach

Rewrite in exponential form:

$$ \log_2\frac{\sqrt{3x+4}}{x-2}= 1 $$ $$ 2^1 = \frac{\sqrt{3x+4}}{x-2} $$ $$ 2(x-2) = \sqrt{3x+4} $$ $$ 2x-4 = \sqrt{3x+4} $$

If you wish, you can solve this using your calculator (treat like a systems of equations and find intersection). Alternatively,

$$ (2x-4)^2 = 3x+4 $$ $$ 4x^2-16x+16 = 3x + 4 $$ $$ 4x^2-19x+12 =0$$ $$ (x-4)(4x-3) = 0 $$ $$ x=4 \text{ and }x=\frac{3}{4} $$

There may be extraneous solutions since we squared both sides. In this case substituting the two values back into the original equation results in only $x=4$ working.