Integrate 7^x * 5^x with respect to x without using the property implying a^x * b^x = (ab)^x. @Calculus1
\[\int\limits{35^x}dx = \frac{35^x}{\ln35}+c\]
For a > 0, \[ \frac{d \ }{dx} a^x = \ln a . a^x \] Thus integrating by parts \[ J = \int 7^x 5^x \ dx = \frac{1}{\ln 7} 7^x 5^x - \frac{\ln 5}{\ln 7} \int 7^x 5^x \ dx \] This last integral is a constant times the original integral J. Now solve algebraically for J.
@sarah, yes exactly. But the question explicitly asks that we don't reduce \(7^x5^x\) to \(35^x\).
@James, is this an instance of when the integral repeats?
If you will, yes.
It's a slightly tortured question and Sarah's approach is the natural one. The good news is you can check your answer against hers.
I don't know if my Algebra is extremely terrible or I've misunderstood your directions, but when I attempted to solve the problem my answer resulted in Ln 7 + Ln 5 = 7^x * 5^x...
What you should find is that \[ (\ln 7 + \ln 5) J = 7^x5^x \] i.e., \[ J = \frac{1}{\ln 7 + \ln 5} 7^x5^x \]
Now check and make sure this is the same as Sarah's (correct) answer.
Sarah's answer is equal to yours. What I don't comprehend is how you use the variable for your answer in your equation.
are you talking about his integrating by parts technique?
i'm not really sure what you are saying
\[ \int\limits 7^x 5^x \ dx = \frac{1}{\ln 7} 7^x 5^x - \frac{\ln 5}{\ln 7} \int\limits 7^x 5^x \ dx \] now we are trying to evaluate \[\int\limits_{}^{}7^x 5^x dx\] do you know how to solve y=a-cy for y? y+cy=a y(1+c)=a y=a/(1+c) so just Let \[J=\int\limits_{}^{}7^x5^x dx\] \[ J = \frac{1}{\ln 7} 7^x 5^x - \frac{\ln 5}{\ln 7} J\] solve this equation for J
so add \[\frac{\ln(5)}{\ln(7)} J\] on both sides
I got it myininaya and James, thanks.
\[\frac{\ln(5)}{\ln(7)}J+J=\frac{1}{\ln(7)} 7^x5^x\] oh ok.... nvm then
just don't forget to put pluc C somewhere
I definitely won't forgot that myininaya!
ok lol
I am sorry James i misread it :(
Join our real-time social learning platform and learn together with your friends!