Which of the following equations is equivalent to 25x = 7

Which of the following equations is equivalent to 25x = 7? A. x = log2(3) + 1 + 9 = 3? B. x = log27(5) C. x = log2(5) + log5 D. x = 2

The Correct Answer and Explanation is:

We are given the equation:

25^x = 7

We are asked to find which of the given choices is equivalent to this equation. To solve this, we need to isolate x using logarithms.

Step 1: Take logarithms on both sides

Take the logarithm base 10 (or natural logarithm) of both sides:log⁡(25x)=log⁡(7)\log(25^x) = \log(7)log(25x)=log(7)

Apply the logarithmic rule:log⁡(ab)=b⋅log⁡(a)\log(a^b) = b \cdot \log(a)log(ab)=b⋅log(a)x⋅log⁡(25)=log⁡(7)x \cdot \log(25) = \log(7)x⋅log(25)=log(7)

Step 2: Solve for x

x=log⁡(7)log⁡(25)x = \frac{\log(7)}{\log(25)}x=log(25)log(7)​

Now, express 25 as 525^252. So:log⁡(25)=log⁡(52)=2⋅log⁡(5)\log(25) = \log(5^2) = 2 \cdot \log(5)log(25)=log(52)=2⋅log(5)

Thus:x=log⁡(7)2⋅log⁡(5)x = \frac{\log(7)}{2 \cdot \log(5)}x=2⋅log(5)log(7)​

This expression is equal to:x=12⋅log⁡(7)log⁡(5)=12⋅log⁡5(7)x = \frac{1}{2} \cdot \frac{\log(7)}{\log(5)} = \frac{1}{2} \cdot \log_5(7)x=21​⋅log(5)log(7)​=21​⋅log5​(7)

This is not directly shown in any of the answer choices, so let’s analyze each:

Evaluate the Choices:

A. x = log₂(3) + 1 + 9 = 3 → This is incorrect and does not relate logically to the original equation.

B. x = log₂₇(5)
This means:27x=527^x = 527x=5

But the original equation is:25x=725^x = 725x=7

These bases and results are different. So this is incorrect.

C. x = log₂(5) + log(5)
This is an invalid expression because log₂(5) and log(5) are not equal or combinable directly. Also, nothing connects it to the equation 25^x = 7. Incorrect.

D. x = 2
Let’s test:252=625≠725^2 = 625 \neq 7252=625=7

So incorrect.

Final Answer:

None of the options are correct. However, if forced to choose the closest match, none correctly simplify to x=log⁡(7)log⁡(25)x = \frac{\log(7)}{\log(25)}x=log(25)log(7)​.

So, the accurate value of x from 25x=725^x = 725x=7 is:x=log⁡(7)log⁡(25)x = \frac{\log(7)}{\log(25)}x=log(25)log(7)​

Or:x=log⁡25(7)x = \log_{25}(7)x=log25​(7)

This is the correct equivalent equation.