Solve log4 32=x+2. Can someone please help me solve this?

- anonymous

Solve log4 32=x+2. Can someone please help me solve this?

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- anonymous

4 is the base of log.

- anonymous

\[\log _{4}32=x+2\]

- myininaya

To solve for x
Subtract 2 on both sides

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## More answers

- anonymous

So, log4 30=x?

- anonymous

Change it to the exponential form. Tell me what you see

- anonymous

I think my way is easier :p try this first ^

- anonymous

4^(x+2)=32
do you know how to get that?

- anonymous

No I don't :/

- anonymous

okay, look at this
http://www.google.ca/imgres?num=10&um=1&hl=en&safe=off&biw=1312&bih=784&tbm=isch&tbnid=U5kYdBmihVLWHM:&imgrefurl=http://people.richland.edu/james/lecture/m116/logs/logs.html&docid=kxHBLrjVjENnqM&imgurl=http://people.richland.edu/james/lecture/m116/logs/log2.gif&w=305&h=298&ei=ckm9T5yVDYnG6AGlmrhR&zoom=1&iact=hc&vpx=180&vpy=140&dur=1829&hovh=222&hovw=227&tx=183&ty=65&sig=112104045567003345819&sqi=2&page=1&tbnh=145&tbnw=148&start=0&ndsp=25&ved=1t:429,r:0,s:0,i:69

- myininaya

\[\log_432=x+2\]
-2 -2
\[\log_432-2=x\]

- anonymous

The whole point of logs, is that they were "made" to "Solve for" an exponent. So if I had an expression that was 2^x=100, the only way to "find x" is to use logarithms. How? You can re-write any exponential equation to look like a logarithm, and vise versa.

- anonymous

To go from log to exponential, you need to write the "answer" as the EXPONENT of the BASE.
4 is the base of the log, so write x+2 as the exponent of 4. 4^(x+2)

- anonymous

and what does that equal? Well you only have one number left; the number you were taking the log of. 32.
This is a basic skill you need to learn :) Writing logs as exponentials, and writiing exponentials as logs.

- myininaya

Now if you want to simplify \[\log_432\]
I would say let g equal that
so we have
\[g=\log_432\]
writing in exponential form we get
\[4^g=32\]
g=3 doesn't work b/c 4(4)(4)=16(4)=64
g=3/2 doesn't work b/c (4)^(3/2)=2^3=8
g=5/2 works b/c (4)^(5/2)=2^5=32
Booyah! :)
So that means
\[\log_4(32)=\frac{5}{2}\]
So what does that mean x equals or simplifies to?

- myininaya

You could just use your calculator though instead of that method right there
sometimes that method will be too difficult to use

- myininaya

A good formula to remember
is the change of base formula
\[\frac{\ln(a)}{\ln(b)}=\log_b(a)\]

- anonymous

\[4^{x+2}=32 \]
So how do you solve this? There's a trick you'll probably have to often use: you need express both numbers as the same base! You can probably tell that 4 is simply 2^2, and 32 is 2^5
That's what I mean by "the same base"
So once you do that, you've got this:
\[2^{2(x+2)}=2^{5}\]
Once you have something looking like that, you can solve for x! You simply "ignore" the "big" 2s, and you get
2(x+2)=5
solve for x now, and you get 0.5, or 1/2 :)

- myininaya

using my way we get
\[\frac{5}{2}-2=x\]

- myininaya

Remember since \[\log_4(32)=\frac{5}{2}\]

- anonymous

The trick to these is:
1. writing it in exponential form (learn to do this!! you'll use it all the time)
2. Expressing both sides of your new (but equilivant to the log expression) equation with the same bases.
3."zooming in" on the exponents, ignoring the "big numbers" (which are now the same) and solving for x like a regular algebra problem.

- anonymous

myininaya i like your solution, it's just that I'm close to the asker's age, I think, and I remember that this is the particular way they want these questions done, even if it is a drag , lol

- anonymous

they=the teachers

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