If $x$ is a real number such that $\log_3 5 = \log_5 (2 + x)$, then which of the following is true?

We have: $\log_3 5 = \log_5 (2 + x)$

Instead of trying to solve this directly (which would be messy), we'll use the fact that both sides are equal to find bounds for $x$.

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Here's the key insight: We need to figure out what range $\log_3 5$ falls into.

Since logarithms are increasing functions, we can compare:

$\log_3 3 = 1$ (because $3^1 = 3$)

$\log_3 9 = 2$ (because $3^2 = 9$)

Since $3 < 5 < 9$, and logarithm is an increasing function:

$\log_3 3 < \log_3 5 < \log_3 9$

$1 < \log_3 5 < 2$

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Since $\log_3 5 = \log_5 (2 + x)$, we can substitute:

$1 < \log_5 (2 + x) < 2$

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This is where many students get confused. When we have $1 < \log_5 (2 + x) < 2$, we can "undo" the logarithm by applying the exponential function with base 5 to all parts.

Remember: If $\log_5 y = z$, then $5^z = y$

So: $1 < \log_5 (2 + x) < 2$ becomes:

$5^1 < 2 + x < 5^2$

$5 < 2 + x < 25$

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Now we just need to isolate $x$:

$5 < 2 + x < 25$

Subtract 2 from all parts:

$5 - 2 < x < 25 - 2$

$3 < x < 23$

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Therefore, $x$ must satisfy: $3 < x < 23$

This approach is brilliant because instead of trying to solve the equation directly, we used the fact that both sides are equal to create bounds. This avoids complex logarithmic calculations and gives us a clear range for $x$.

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Alternative method:

We are given: $\log_3 5 = \log_5 (2+x)$

Let $\log_3 5 = \log_5 (2+x) = k.$

From the first equation, $3^k = 5.$

From the second equation, $5^k = 2+x.$

Since $k = \log_3 5,$

we substitute into the second equation: $2+x = 5^{\log_3 5}.$

Using the identity: $a^{\log_b c} = c^{\log_b a},$

we get $2+x = 3^{(\log_3 5)^2}.$

This is the exact value.

Now, Using $\log_3 5 = \frac{\ln 5}{\ln 3} \approx 1.465,$

we have $2+x = 5^{1.465} \approx 10.56.$

Therefore, $x \approx 10.56 - 2 = 8.56.$