Question

solve the following equations: 2^(-100x)=(0.5)^(X-4) and 4^x * (1/2)^(3-2x)=8 * (2^x)^2

Answer 1

\[2^{(-100x)}=(0.5)^{(X-4)}\]
do you know how to use logarithms on this?

Answer 2

i know to get an equivalent base and that is what i don't know how to get.

Answer 3

\[a^B=c^F~~~~~~~->~~~~~~\log(a^B)=Log(c^F)\]do this process and use the\[\log(a^b)=b~\log~a\]

Answer 4

is there a way without logs, that is a future lesson. not for this one.

Answer 5

yes. write 0.5 as 1/2 = 2^(-1)
The bases will be the same on both sides and therefore the exponents will be the same.

Answer 6

Man!! Haven't though of that!

Answer 7

thank you

Answer 8

you are welcome.