which of the following shows the true solution to the logarithmic equation 3log₂(2x)=3?\no x = - 1\no x =…

Explanation:

Step1: Isolate the logarithm

Divide both sides of the equation $3\log_2(2x)=32$ by 3. $\log_2(2x)=\frac{32}{3}$

Step2: Convert to exponential form

Using the property $\log_a b = c$ is equivalent to $a^c = b$, we have $2^{\frac{32}{3}}=2x$.

Step3: Solve for x

Divide both sides by 2: $x = \frac{2^{\frac{32}{3}}}{2}=2^{\frac{32}{3}- 1}=2^{\frac{29}{3}}$. But it seems there is a typo in the original equation as the given options don't match with the correct solution. If the equation was $3\log_2(2x) = 3$, then:

Step1 (for corrected equation): Isolate the logarithm

Divide both sides by 3, we get $\log_2(2x)=1$.

Step2 (for corrected equation): Convert to exponential form

By the property $\log_a b = c\Rightarrow a^c = b$, we have $2^1 = 2x$.

Step3 (for corrected equation): Solve for x

Divide both sides by 2, $x = 1$.

Answer:

$x = 1$ (assuming the corrected - equation situation)