Solving an exponential equation by factoring powers

Key takeaway: This exponential equation becomes simple once the common factor $2^x$ is pulled out. After that, the problem reduces to a linear equation in an exponential expression.

Start by factoring the common power

The equation is

$2^{x+2}-2^x=48.$

Rewrite the first term:

$2^{x+2}=2^x\cdot 2^2=4\cdot 2^x.$

So the equation becomes

$4\cdot 2^x-2^x=48.$

Factor out $2^x$:

$2^x(4-1)=48.$

That gives

$3\cdot 2^x=48.$

Solve for $x$

Divide both sides by $3$:

$2^x=16.$

Since $16=2^4$, we have

$x=4.$

Check the solution

Substitute $x=4$ into the original equation:

$2^{4+2}-2^4=2^6-2^4=64-16=48.$

The check works, so the solution is

$\boxed{x=4}.$

Why factoring works here

Both terms contain a power of $2$ with the same base. That makes factoring the shared $2^x$ the fastest method. Once the exponential part is isolated, the equation turns into a basic power comparison.

This is a standard pattern in exponential equations: rewrite each term with the same base, factor the common exponential expression, solve the resulting simpler equation, and then verify the result.

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Pitfalls the pros know 👇 A common mistake is trying to divide both sides by $2^x$ too early without first noticing that $2^x$ is a common factor on the left. While that can work here, it is easy to lose track of the structure and make algebra errors. Another issue is forgetting the check step. Exponential equations can sometimes produce extraneous-looking results if you manipulate them poorly, so substituting back into the original equation is good practice. Also, be careful not to treat $2^{x+2}$ as $2^x+2^2$; exponents do not distribute over addition.

What if the problem changes? If the equation were $2^{x+3}-2^{x+1}=48$, the same method would apply: factor out the smallest power of 2, giving $2^{x+1}(4-1)=48$, so $2^{x+1}=16$ and $x=3$. If the right-hand side were not a power of 2, for example $2^{x+2}-2^x=50$, the same factoring step would lead to $3\cdot 2^x=50$, so $x=\log_2(50/3)$ instead of an integer.

Tags: common factor in exponents, power of the same base, exponential equation check