3^(7x-1) = 5^(3x+2) I've tried to take the natural log of both sides, and pull x out of the exponent, but I can't seem to get the right answer.
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Originally Posted by Velvet Love 3^(7x-1) = 5^(3x+2) I've tried to take the natural log of both sides, and pull x out of the exponent, but I can't seem to get the right answer. $\displaystyle \ln{3^{(7x-1)}}=\ln{5^{(3x+2)}}$ $\displaystyle \Rightarrow(7x-1)\ln{3}=(3x+2)\ln{5}$ Can you take it from here?
That's what I've gotten it down to. Then I end up with ((ln(5)/ln(3))+3)/4, which is wrong. I'm rusty at pre-cal.
Originally Posted by Velvet Love That's what I've gotten it down to. Then I end up with ((ln(5)/ln(3))+3)/4, which is wrong. I'm rusty at pre-cal. Distribute: $\displaystyle (7\ln3)x-\ln{3}=(3\ln{5})x+2\ln{5}$ Isolate x: $\displaystyle (7\ln3)x-(3\ln{5})x=2\ln{5}+\ln3$ $\displaystyle x(7\ln{3}-3\ln{5})=2\ln{5}+\ln3$ And now divide...
Thank you alot. That helped a lot. I'm glad I learn best by example
I worked out 1.51..... is that correct?
Originally Posted by sanddollar I worked out 1.51..... is that correct? yes, if your answer is 3s.f.
Originally Posted by Velvet Love I end up with ((ln(5)/ln(3))+3)/4, which is wrong. It'll be a lot easier to figure out where you might be going wrong if you show your steps.
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