Question

Will fan if help with logs!

solve for x. 5^x-1 = 2

Answer 1

\[\Large 5^{x-1} = 2\]Take logs of both sides \[\Large \log 5^{x-1} = \log 2\] \[\Large (x-1)\log 5 = \log 2\]

Answer 2

Shouldn't 5 be the base? Same with 2

Answer 3

You have to take the same base of both sides, can't take diff. bases. And log base 5 is not on a calculator.

When taking logs of both sides, you almost always want to use log (base 10) or ln.

Answer 4

but since x=log(base)y is the same as y=b^x would 5 not be the base?

Answer 5

Again, because log base 5 is not on a calculator, and you can't do log base 5 of 2 without one. You don't need to use that method here.

Answer 6

ok, so we are basically just implementing logs into the equation right?

Answer 7

Taking logs of both sides, in any equation similar to this one, will ALWAYS work.

Answer 8

ok thank you

Answer 9

Can you solve it from \[\Large (x-1)\log 5 = \log 2\]

Answer 10

this ok?