Question

which of the following shows the true solution to the logarithmic equation $3\log_{2}(2x)=3$?$x=-1LXB0x=-1$ and $x=1$$x=0$, $x=-1$, and $x=1$

Explanation:

Step1: Isolate the logarithm term

Divide both sides by 3:
$$\log_{2}(2x) = \frac{3}{3} = 1$$

Step2: Convert to exponential form

Use $\log_b(a)=c \implies b^c=a$:
$$2^1 = 2x$$

Step3: Solve for x

Divide both sides by 2:
$$x = \frac{2}{2} = 1$$

Step4: Verify domain

Logarithm argument $2x>0 \implies x>0$. $x=-1$ is invalid as it makes $2x=-2<0$.

Answer:

x=1