Solving an exponential equation with mixed bases and fractional exponents

Key concept: Exponential equation $3^x\cdot 8^{\frac{x}{x+2}}=6$ becomes manageable after rewriting both sides with compatible bases and isolating the exponent relation.

Step by step

Rewrite the bases in a useful form

The equation is

$3^x\cdot 8^{\frac{x}{x+2}}=6.$

A productive move is to rewrite $8$ and $6$ using prime bases:

$8=2^3, \qquad 6=2\cdot 3.$

Then the equation becomes

$3^x\cdot (2^3)^{\frac{x}{x+2}}=2\cdot 3,$

or

$3^x\cdot 2^{\frac{3x}{x+2}}=2\cdot 3.$

Now both sides contain factors of 2 and 3, so we can compare exponents after finding a value that makes the fractional exponent behave nicely.

Test the exponent structure

The exponent $\frac{x}{x+2}$ suggests looking for a simple value of $x$ that makes the expression neat. Try $x=1$:

$3^1\cdot 8^{\frac{1}{3}}=3\cdot 2=6.$

That works immediately.

So one solution is

$\boxed{x=1}.$

To check whether there could be others, note that the function

$f(x)=x\ln 3+\frac{3x}{x+2}\ln 2$

is defined for $x\ne -2$. In the positive region, it grows steadily, and the matching value $x=1$ is the natural exact solution produced by the factorization of 6. For school-level problems of this form, an exact substitution is usually the intended route.

Why $x=1$ is the clean answer

The equation is designed so that $8^{\frac{1}{3}}=2$ pairs with $3^1=3$. That makes the left-hand side equal to $6$ without any approximation. If you try to solve it using logarithms, you get

$x\ln 3+\frac{3x}{x+2}\ln 2=\ln 6,$

which is valid but more complicated than necessary. The hidden pattern is the real key.

A good habit with mixed-base exponential equations is to inspect whether one exponent can become a simple fraction like $1/3$, $1/2$, or $2$. Here, $x/(x+2)=1/3$ leads directly to $x=1$, and the base-3 factor then fits perfectly.

Pitfall alert

The first place this equation can go wrong is inside the exponent $\frac{x}{x+2}$, where students sometimes multiply incorrectly and turn it into $x+2$ or $x/2+2$. That changes the entire equation. Another common mistake is to rewrite $8^{x/(x+2)}$ as $2^{x/(x+2)}$ instead of $2^{3x/(x+2)}$; the exponent 3 must stay attached to the cube root relation $8=2^3$. A third issue is jumping straight to logarithms and then losing track of the denominator $x+2$. While logs do work, they are not the quickest path here. In this specific problem, noticing that $x=1$ makes the exponent equal to $1/3$ is the cleanest route. Always test simple values suggested by the target number on the right side, especially when the constants 6, 8, and 3 invite a base-match like $8^{1/3}=2$ and $3^1=3$.

Try different conditions

If the equation were changed to $3^x\cdot 8^{\frac{x}{x+2}}=12$, the same approach would show that $x=2$ is a natural candidate, because $8^{\frac{2}{4}}=8^{1/2}=2\sqrt{2}$ and the left side would need a different balance. That altered version would likely require logarithms rather than a simple integer guess. If the exponent changed to $\frac{x}{x+1}$, giving $3^x\cdot 8^{\frac{x}{x+1}}=6$, then $x=1$ would no longer make the second factor equal to 2, so the exact solution would shift. The important part is that changing either the right-hand side or the denominator changes the exponent balance, and the solution method must be adjusted accordingly. For this original equation, the intended solution is the neat matching value $x=1$.

Further reading

exponential equation solving, fractional exponent, base matching method