Question

What are the rules for solving for x?

8^x = (1/32)^(x-2)

Answer 1

\[8^x = (\frac{ 1 }{ 32 })^{x-2}\]

Answer 2

yuck ok so, first question, what can you do to bring the exponents down?

Answer 3

You have two choices, factor or use logarithms.

Answer 4

(I personally prefer logs, and it is not nearly as yucky as it was upon first inspection)

Answer 5

Factoring would lead to the realization that \(8=2^3\) and \(1/32 = 2^{-5}\).

Answer 6

\(\LARGE\color{black}{ 8^x=(\frac{1}{32})^{x-2} }\)

\(\LARGE\color{black}{ 8^x=(\frac{1}{2^5})^{x-2} }\)

\(\LARGE\color{black}{ 8^x=(2^{-5})^{x-2} }\)

\(\LARGE\color{black}{ (2^3)^x=(2^{-5})^{x-2} }\)

\(\LARGE\color{black}{ (2)^{3x}=(2^{-5})^{x-2} }\)

\(\LARGE\color{black}{ (2)^{3x}=(2^{})^{-5(x-2)} }\)

Answer 7

hope you can take it from here.

Answer 8

Yep Solomon Thanks!

Answer 9

I don't usually notice that type of stuff, nice catch wio, I need a more accross the board method

Answer 10

When working with logs, you generally want to factor anyway.

Answer 11

sort of, instead of having to notice the ways you can rewrite two, you can take a log of both sides, then by properties of logs, you can bring down the exponents and solve for x

Answer 12

Ohh ok so Solomon's method would be my number one choice?

Answer 13

@OwdinMW Maybe you should do your own homework. Maybe you'd learn something that way.

Answer 14

it's up to personal preference essentially

Answer 15

\[\frac{1}{a^m} = a^{-m} \\ (a^m)^n = a^{m \times n} \\ If \ \ \ a^m = a^n, \implies m = n\]

Answer 16

Alright :) thank you!

Answer 17

Have you studied logarithms @OdinMW ?

Answer 18

Yes. I'm just getting into them.

Answer 19

If you are just getting into them, chances are you should do the problem both ways to ensure you recieve credit

Answer 20

\[\log_{b}\frac{ m }{ n } = \log_{b} m - \log_{b} n\]

Answer 21

have you learned about how to bring down exponents with logs?

Answer 22

@FibonacciChick666 yes

Answer 23

First thing you want to do here is change the fraction into a negative exponent. \[
8^x=32^{2-x}
\]

Answer 24

i concur

Answer 25

Then you would use logs.

Answer 26

then what would happen if you took log base 8 of both sides?

Answer 27

Yes @wio thats something I should have thought of.

Answer 28

You can keep the base to be whatever you want for the beginning.

Answer 29

or ln, either works

Answer 30

So the log of both sides gives: \[
x\log 8=(2-x)\log32
\]

Answer 31

personally, I prefer base 8 here because it makes less work later

Answer 32

and I think you can manage from there odin

Answer 33

8

Answer 34

Yes

Answer 35

Whoops, I wrote down the wrong thing.

Answer 36

i got, more along the lines of x=2log32-xlog32

Answer 37

the log in this case was base 8

Answer 38

Im thinking its getting a little too much haha. I would like to know how to solve it with logs but I think Ill just use Solomon's method...

Answer 39

loooooossssseeeeerrrrrrrrrsssssssssss

Answer 40

Hmmm:\[
x\log 8=(2-x)\log32\\
x\log8=2\log32-x\log32\\
x\log8+x\log 32 = 2\log 32\\
x(\log8+\log32)=2\log32\\
x = \frac{2\log 32}{\log8+\log32}
\]

Answer 41

x=2log32-xlog32< this is just a two step problem from here, it's not that bad, I swear :)

Answer 42

or what wio did

Answer 43

Change it to base \(2\), and it becomes: \[
x = \frac{2(5)}{(3)+(5)}
\]

Answer 44

If you had \[
\log_8(32)
\]You can always convert this to another base \(c\) using the property: \[
\log_ab = \frac{\log_cb}{\log_ca}
\]

Answer 45

In general, you either want your base to be a prime number, or you just want to use the natural log \(\ln\).

Answer 46

\[x=32\] basically...right?