Evaluate the integral: integral 5^x 3^-x dx
Evaluate the integral: $$ \int 5^x 3^{-x} \, dx. $$
Explanation of Method
A product of exponentials with the same variable exponent but different bases, $a^x b^{-x}$, combines into a single exponential $(a/b)^x$ using the exponent rule $a^m b^{-m} = (a/b)^m$. Once combined, the integral follows directly from the general exponential antiderivative formula.
- Combine the product of exponentials into a single base raised to the $x$.
- Apply the general exponential integration formula.
- Write the final antiderivative and add the constant of integration.
Step 1 · Simplify the integrand
The integrand is $5^x 3^{-x}$. Using the property of exponents, $a^m b^{-m} = (a/b)^m$, we can combine the terms:
$$ 5^x 3^{-x} = \left(\frac{5}{3}\right)^x. $$
Thus, the integral becomes:
$$ \int 5^x 3^{-x}\,dx = \int \left(\frac{5}{3}\right)^x dx. $$
Step 2 · General formula for exponential integrals
The integral of $a^x$ with respect to $x$ is:
$$ \int a^x\,dx = \frac{a^x}{\ln(a)} + C, \quad \text{for } a>0,\ a\neq 1. $$
Here, $a = \dfrac{5}{3}$. Applying this formula, we get:
$$ \int \left(\frac{5}{3}\right)^x dx = \frac{\left(\frac{5}{3}\right)^x}{\ln\left(\frac{5}{3}\right)} + C. $$
Step 3 · Final expression
Substituting back the simplified result, the integral evaluates to:
$$ \int 5^x 3^{-x}\,dx = \frac{\left(\frac{5}{3}\right)^x}{\ln\left(\frac{5}{3}\right)} + C, $$
where $C$ is the constant of integration.
Summary Table
| Original term | Combined base | Antiderivative |
|---|---|---|
| $5^x 3^{-x}$ | $(5/3)^x$ | $\dfrac{(5/3)^x}{\ln(5/3)}$ |
Additional Notes
- Since $5/3 > 1$, $\ln(5/3) > 0$, so the sign of the antiderivative matches the sign convention used for bases greater than $1$.
- The integrand is defined for all real $x$, since exponential functions with positive base are defined everywhere.
- The result can be checked by differentiating $\dfrac{(5/3)^x}{\ln(5/3)}$ and confirming it returns the original integrand.
Result
The solution to the integral is: