Expressing a power product as one base three exponent

Key concept: The expression k(x)=(9^(x+4)·3^(4x-1))/3^x becomes manageable once every factor is written with base 3 [1].

Step by step

Convert everything to base 3

The expression $k(x)=\dfrac{(9^{x+4})\cdot 3^{4x-1}}{3^x}$ is built from powers of 9 and 3, so the first move is to rewrite $9$ as $3^2$. That gives

$9^{x+4}=(3^2)^{x+4}=3^{2(x+4)}=3^{2x+8}.$

Now substitute this into the original expression:

$k(x)=\frac{3^{2x+8}\cdot 3^{4x-1}}{3^x}.$

Combine exponents carefully

The numerator has the same base, so multiply by adding exponents:

$3^{2x+8}\cdot 3^{4x-1}=3^{(2x+8)+(4x-1)}=3^{6x+7}.$

Then divide by $3^x$, which means subtract exponents:

$\frac{3^{6x+7}}{3^x}=3^{(6x+7)-x}=3^{5x+7}.$

So the expression is already in the requested form $3^{(ax+b)}$, with $a=5$ and $b=7$.

Final answer

$\boxed{k(x)=3^{5x+7}}.$

This matches the required pattern exactly:

$3^{ax+b}=3^{5x+7}.$

Why the exponent rules work here

When bases are identical, exponent laws are reliable and fast:

• $a^m\cdot a^n=a^{m+n}$
• $a^m/a^n=a^{m-n}$
• $(a^m)^n=a^{mn}$

The only nontrivial step in this problem is recognizing that $9=3^2$. Once that conversion is made, the rest is exponent bookkeeping.

A good habit is to rewrite the entire expression in one base before combining anything. That reduces the risk of mixing base 9 and base 3 in the same line, which often leads to incorrect exponent arithmetic.

Common mistake to avoid

A common error is to treat $9^{x+4}$ as $9x+4$ or to multiply the exponent by 2 incorrectly as $3^{2x+4}$. The exponent $x+4$ multiplies by 2 because $(3^2)^{x+4}=3^{2(x+4)}$, so both terms inside the parentheses are multiplied. Another mistake is forgetting that division subtracts exponents. If you leave the denominator in place as another factor of 3, the final form will not match the requested single-exponent expression.

If you keep track of base conversion first, then addition and subtraction of exponents, this kind of problem becomes routine and very fast to check.

Pitfall alert

The step that usually causes trouble is converting $9^{x+4}$ into a power of 3, because the exponent $x+4$ must distribute across the 2 coming from $9=3^2$. Students often write $3^{2x+4}$, but that drops the factor of 2 on the constant 4 and gives the wrong result. The correct expansion is $3^{2(x+4)}=3^{2x+8}$. Another place mistakes appear is in the division by $3^x$. Since division subtracts exponents, you must reduce $3^{6x+7}/3^x$ to $3^{5x+7}$, not $3^{6x-7}$ or $3^{x+7}$. If you are checking your work, convert the entire answer back into a product of powers and verify that the exponent bookkeeping matches every factor from the original expression. The form $3^{ax+b}$ is only correct if there is exactly one exponent on base 3 at the end.

Try different conditions

If the problem were changed to $k(x)=\dfrac{(27^{x+4})\cdot 3^{4x-1}}{3^x}$, then the rewritten base would be $27=3^3$ instead of $9=3^2$. The new expression becomes $k(x)=\dfrac{3^{3(x+4)}\cdot 3^{4x-1}}{3^x}=\dfrac{3^{3x+12}\cdot 3^{4x-1}}{3^x}$. Combining the numerator gives $3^{7x+11}$, and dividing by $3^x$ gives $3^{6x+11}$. So the final answer would be $3^{6x+11}$. This variant shows that the structure of the solution stays the same, but the coefficient of $x$ and the constant term both change depending on how many times the exponent from the base conversion multiplies the bracket. If the base were instead $81$, the same pattern would continue because $81=3^4$.

Further reading

exponent rules with same base, rewriting powers of 9, single base exponential form