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Solve for x. 7^x=e^(x+5)

Mathematics
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If you take the ln on both sides, you can use the rule: \[\ln a^b = b \cdot \ln a\]so:\[\ln 7^x=\ln e^{x+5}\]By the rule, this is equal to\[x \cdot \ln7 = (x+5) \ln e\] Of course, you know that ln e = 1, so you can solve for x now...
as the matter of fact, I was able to come up with that but im confused when it comes x. so, the answer should be x=5/In(7)?
xln7=x+5, so xln7 - x = 5. Factor out x: x(ln7 - 1)=5. Divide by ln7 -1:\[x=\frac{ 5 }{ \ln7 -1 }\]

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wow, that makes sence! thank you so much! :)
May I ask another question? im stuck with this math question. 5=55(1.3)^x. for x using logs to slove the equation.

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