Question

I kinda forgot how to do this, but I know it has to do with logarithms

17^(5x) = 8^(x+5)

Solving for x.

Answer 1

\(\Large 17^{5x} = 8^{x+5}\)

Answer 2

If I take the log on both sides

\(\Large 5x log(17) = (x+5)log8\)

I don't really remember how to isolate x :P

Answer 3

This is supposed to be easy, I remember solving questions like this, but I guess I'm getting rusty xD

Answer 4

distribute the `log(8)` through and then get all the x terms to one side

Answer 5

\(\Large 5x log(17) = xlog(8) + 5log(8)\)

\(\Large 5x log(17) -xlog(8)= 5log(8)\)

\(\Large x(5log(17) -log(8))= 5log(8)\)

\(\Large x = \frac{5log(8)}{5log(17) -log(8)}\)

Answer 6

you nailed it

Answer 7

It was just that distributing that log(8) that had me stuck >.<

Thanks, Jim! You're the best! :)

Answer 8

log(8) is just another constant, like "3" or "√2". Hope that helps in the future!

Answer 9

mhmmm

log(8) = log(2^3) = 3log(2)

Answer 10

I blame the obsolescence of slide rules for everyone's struggles with logarithmic properties.

Answer 11

lol - that also confuses me sometimes !

Answer 12

- distributing the log I mean.

Answer 13

Yeah, I had a similar question for this assessment I had to take to show that I would be ready to start Calc in fall for Univ. But I was like this question has me stumped, and I knew how to solve them ages ago.

Answer 14

You really should review algebra in this case. Also your trig identities. Forget to do it, and you'll regret getting those annoying B's on your tests ;)

Answer 15

Yes, however human error can't be eliminated.

Answer 16

That is true. It's nice though in calculus because generally you can take the inverse operation to check your answer. I easily got from a high 80 or low 90 to a high 90 just by taking time after a test to check my work. The more different the checking method is from the original, the more errors I tend to catch.

Answer 17

Good.