I'm doing logarithms in Math 12 and I'm a bit confused I was wondering how I could solve the following problem:
log2 (x+1) = 1+ log4(3x-2)
I'll kick it off for you.
$\displaystyle \log_2 (x+1) = 1+ \log_4(3x-2) $
$\displaystyle \log_2 (x+1) = \log_44+ \log_4(3x-2) $
$\displaystyle \log_2 (x+1) = \log_4(4(3x-2)) $
$\displaystyle \log_2 (x+1) = \log_4(12x-8)) $
now you need to change bases to either 4 or 2.
Just in case there were others who did not understand how to proceed from what pickslides had written, this is how one would finish the problem:
log_2 (x+1) = 1 + log_4 (3x-2)
4^log_2 (x+1) = 12x-8 (convert the above to exponential form)
2^log_2 (x+1)^2 = 12x-8 (convert the 4 into 2^2 and then use the property of logarithms to move the latter 2 (the one after the carrot) from front of the log sign so that it can square (x+1))
(x+1)^2 = 12x-8 (cancel out the other 2_log 2 based on the property of logarithms)
x^2 + 2x +1 = 12x-8 (solve the quadratic equation)
x^2 -10x + 9 = 0
x=1, x=9
Since neither solution is negative, both solutions are valid and will work. I have checked using a calculator to make sure.
I would have done the first problem is a slightly different way. From $\displaystyle log_2 (x+1) = 1+ log_4(3x-2)$, subtract $\displaystyle log_4(3x-2)$ from both sides to get $\displaystyle log_2(x+1)- log_4(3x-2)= 1$.
Now change one of the bases: $\displaystyle log_4(x)= y$ means $\displaystyle x= 4^y= (2^2)^y= 2^{2y}$ and from that $\displaystyle 2y= log_2(x)$. Thus, $\displaystyle log_4(3x-2)= \frac{1}{2}log_2(3x-2)= log_2(3x-2)^{1/2}$ and the equation becomes $\displaystyle log_2(x+1)- log_2(3x-2)^{1/2}$$\displaystyle = log_2\left(\frac{x+1}{(3x-2)^{1/2}}= 1$.
And that gives $\displaystyle \frac{x+1}{(3x-2)^{1/2}}= 2$. Now you can write that as $\displaystyle x+1= 2(3x-2)^{1/2}$, square both sides and finish as before.
As for $\displaystyle log_5 (log_x(log3 27)) = 1$ which is, indeed, the same as $\displaystyle log_5 (log_x(3)) = 1$, I would recommend that now you use the fact that if $\displaystyle log_5(A)= 1$, then $\displaystyle A= 5^1= 5$.
That is, you now have $\displaystyle log_x(3)= 5$ and that is the same as saying $\displaystyle 3= 5^x$. Solve that for x by using whatever logarithm is handy, either "common" or "natural. $\displaystyle log(3)= log(5^x)= xlog(5)$ so $\displaystyle x=\frac{log(3)}{log(5)}$. That cannot be done "by hand". Use either common log or natural log, whichever is easier to do on your calculator.
Someone correct me if I'm wrong, but this is what I'm getting for that second one:
$\displaystyle \log_5 (\log_x (\log_3 27)) = 1$
$\displaystyle \log_5 (\log_x 3) = 1$
$\displaystyle \log_5 (\frac {\log_5 3}{\log_5 x}) = 1$
$\displaystyle \log_5 (\log_5 3) - \log_5 (\log_5 x) = 1$
$\displaystyle \log_5(\log_5 3) = 1 + \log_5 (\log_5 x)$
$\displaystyle \log_5(\log_5 3) = \log_5 5 + \log_5 (\log_5 x)$
$\displaystyle \log_5(\log_5 3) = \log_5 5(\log_5 x)$
$\displaystyle \log_5(\log_5 3) = \log_5 (\log_5 x^5)$
$\displaystyle \log_5 3 = \log_5 x^5$
$\displaystyle 3 = x^5$
$\displaystyle x = \sqrt[5] {3}$