anonymous
  • anonymous
Solve log4 32=x+2. Can someone please help me solve this?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
4 is the base of log.
anonymous
  • anonymous
\[\log _{4}32=x+2\]
myininaya
  • myininaya
To solve for x Subtract 2 on both sides

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anonymous
  • anonymous
So, log4 30=x?
anonymous
  • anonymous
Change it to the exponential form. Tell me what you see
anonymous
  • anonymous
I think my way is easier :p try this first ^
anonymous
  • anonymous
4^(x+2)=32 do you know how to get that?
anonymous
  • anonymous
No I don't :/
anonymous
  • 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
  • myininaya
\[\log_432=x+2\] -2 -2 \[\log_432-2=x\]
anonymous
  • 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
  • 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
  • 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
  • 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
  • myininaya
You could just use your calculator though instead of that method right there sometimes that method will be too difficult to use
myininaya
  • myininaya
A good formula to remember is the change of base formula \[\frac{\ln(a)}{\ln(b)}=\log_b(a)\]
anonymous
  • 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
  • myininaya
using my way we get \[\frac{5}{2}-2=x\]
myininaya
  • myininaya
Remember since \[\log_4(32)=\frac{5}{2}\]
anonymous
  • 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
  • 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
  • anonymous
they=the teachers

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