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RacingBabee21 · Feb 11, 2007, 04:18 PM

did I solve this righ?

log 2 + 3 log (X) = log 8
2 2 2

log 2Xraised to the 3rd = log 8
2 2
2X raised to the 3rd = 8
x raised to the 3rd = 8
x= squared root of 4

can anyone tell me if I did this right?

asterisk_man · Feb 11, 2007, 05:55 PM

If I read what you wrote correctly I think your problem is to solve for X given this equation:
$\log_2 2 + 3 \log_2 X = \log_2 8$

The following identities will help:
$\begin{eqnarray} (1)&\log_b m + \log_b n &=& \log_b \left(m*n\right)\\ (2)&n \log_b m &=& \log_b \left(m^n\right)\\ (3)&b^{\log_b m}&=&m\\ \end{eqnarray} $

Now we can begin to solve:
$\log_2 2 + 3 \log_2 X = \log_2 8\\ \text{using identity (2):}\\ \log_2 2 + \log_2 X^3 = \log_2 8\\ \text{using identity (1):}\\ \log_2 {2*X^3}=\log_2 8\\ \text{make each side a the power of 2}\\ 2^{\log_2 {2*X^3}}=2^{\log_2 8}\\ \text{using identity (3):}\\ 2*X^3=8\\ \text{then solve for X:}\\ X^3=4\\ X=4^{\left(\frac 1 3\right)}\approx 1.5874\\ $

Comparing this to what you did it looks like you worked the problem correctly. The only concern I have is your final answer:

x= squared root of 4

The answer is:
x=cube root of 4
not square root of 4.
I suspect that's what you meant to write since you got everything else correct up to that point so good work!

Let us know if you have any additional questions.