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

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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.