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dmartinez130
 2 years ago
Best ResponseYou've already chosen the best response.04 is the base of log.

dmartinez130
 2 years ago
Best ResponseYou've already chosen the best response.0\[\log _{4}32=x+2\]

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.0To solve for x Subtract 2 on both sides

Ravus
 2 years ago
Best ResponseYou've already chosen the best response.1Change it to the exponential form. Tell me what you see

Ravus
 2 years ago
Best ResponseYou've already chosen the best response.1I think my way is easier :p try this first ^

Ravus
 2 years ago
Best ResponseYou've already chosen the best response.14^(x+2)=32 do you know how to get that?

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.0\[\log_432=x+2\] 2 2 \[\log_4322=x\]

Ravus
 2 years ago
Best ResponseYou've already chosen the best response.1The 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 rewrite any exponential equation to look like a logarithm, and vise versa.

Ravus
 2 years ago
Best ResponseYou've already chosen the best response.1To 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)

Ravus
 2 years ago
Best ResponseYou've already chosen the best response.1and 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
 2 years ago
Best ResponseYou've already chosen the best response.0Now 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
 2 years ago
Best ResponseYou've already chosen the best response.0You could just use your calculator though instead of that method right there sometimes that method will be too difficult to use

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.0A good formula to remember is the change of base formula \[\frac{\ln(a)}{\ln(b)}=\log_b(a)\]

Ravus
 2 years ago
Best ResponseYou've already chosen the best response.1\[4^{x+2}=32 \] 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
 2 years ago
Best ResponseYou've already chosen the best response.0using my way we get \[\frac{5}{2}2=x\]

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.0Remember since \[\log_4(32)=\frac{5}{2}\]

Ravus
 2 years ago
Best ResponseYou've already chosen the best response.1The 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.

Ravus
 2 years ago
Best ResponseYou've already chosen the best response.1myininaya 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
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