My Math Forum Help with a few log equations

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 June 8th, 2016, 10:18 AM #1 Newbie   Joined: Jun 2016 From: U.Sa Posts: 1 Thanks: 0 Help with a few log equations A few problems that I'm a little stumped on. A) e^x + 6e^-x - 7 =0 What I tried to do here is move the 6e^-x to the right and convert to natural log. So. ln(e^x-7)=ln (-6e^-x) or ln -6 + -x on the right. Since ln e to an exponent is the exponent. After that I'm stumped. I assume I can make the left side x ln (e-7) = ln -6 + -x but I'm lost after that. B) log base 3 (x-2) + log base 3 (2x-3) = 5 log base 3 (x) I subtracted the right and made it into log base 3 (x-2)(2x-3)/x^5 but that's it.
 June 8th, 2016, 10:39 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,656 Thanks: 2634 Math Focus: Mainly analysis and algebra A) Multiply the original equation through by $\mathrm e^{x}$ and note that $\mathrm e^{2x}=(\mathrm e^x)^2$ B) From where you are use $3^{\log_3 y} = y$ Last edited by v8archie; June 8th, 2016 at 10:41 AM.
June 10th, 2016, 06:00 AM   #3
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Quote:
 Originally Posted by Sudowoodo A few problems that I'm a little stumped on. A) e^x + 6e^-x - 7 =0
Multiply through by e^x to get e^(2x)+ 6- 7e^x= (e^x)^2- 7e^x+ 6= 0. Now let y=e^x. Can you solve y^2- 7y+ 6= 0?

Quote:
 What I tried to do here is move the 6e^-x to the right and convert to natural log. So. ln(e^x-7)=ln (-6e^-x) or ln -6 + -x on the right. Since ln e to an exponent is the exponent. After that I'm stumped. I assume I can make the left side x ln (e-7) = ln -6 + -x but I'm lost after that. B) log base 3 (x-2) + log base 3 (2x-3) = 5 log base 3 (x) I subtracted the right and made it into log base 3 (x-2)(2x-3)/x^5 but that's it.
I hope you mean you changed it to log_3((x-2)(2x-3)/x^5)= 0. So what number has log, base 3, equal to 0? (What is 3^0?)

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