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Jonathan D asked:


I have two concerns. (Bare in mind this is Math 12 Adv. and I have not learned Natural Logs, and didn’t touch much on Algebra)
1. 2^(x-1) = 23(6)^3x

I have done this two ways.. both with different answers.
Way 1: 0.086957^(x-1) = 6^3x Now right from the bat I am not certain I can divide the base by 23..
log(0.086957)^(x-1) = log6^3x
(x-1)log0.086957 = 3x(log6)
xlog0.086957 - log0.086957 = 3xlog6
xlog0.086957 - 3xlog6 = log0.086957
x(log0.086957 - 3log6) = log0.086957
x= log0.086957/ (log0.086957 - 3log6)
x= 0.3124

OR
Way 2: 0.086957^(x-1) = 6^3x
log(0.086957)^(x-1) = log6^3x
(x-1)log0.086957 = 3x(log6)
*Divide both sides by log(0.086957)*
x-1 = 3x (-0.7336)
x-1 = 2.2008x
-1 = 1.2008x
x= -1.2

I don’t know which is right.. or if neither of them are.

Question two: For
log*base 3*X + log*base 2* 5 = log*base 7* 12
I just trial and errored to get
log*base 3*X + 2.32 = 1.277
x = 0.3179
Is that the correct way to solve for x? Sorry for the untidiness.
Okay, I worked it out so I now have (x-1)log(2) = log23 + xlog6 but I have no idea where to go from there.. And I cannot use Ln since we didn’t learn them in class yet, and he doesn’t want us using them, even if we can.

Vivian