# Math Help - Logarithms

1. ## Logarithms

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)

2. I'll kick it off for you.

$\log_2 (x+1) = 1+ \log_4(3x-2)$

$\log_2 (x+1) = \log_44+ \log_4(3x-2)$

$\log_2 (x+1) = \log_4(4(3x-2))$

$\log_2 (x+1) = \log_4(12x-8))$

now you need to change bases to either 4 or 2.

3. Okay I get it now. Thankyou very much

4. ## finishing the problem from above

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.

5. would I use the same method for this equation?

log5 (logx(log3 27)) = 1
log5 ( logx(3) = 1

I know I have to change the base to 5 Im not sure how though, any help is great thankyou

6. I would have done the first problem is a slightly different way. From $log_2 (x+1) = 1+ log_4(3x-2)$, subtract $log_4(3x-2)$ from both sides to get $log_2(x+1)- log_4(3x-2)= 1$.
Now change one of the bases: $log_4(x)= y$ means $x= 4^y= (2^2)^y= 2^{2y}$ and from that $2y= log_2(x)$. Thus, $log_4(3x-2)= \frac{1}{2}log_2(3x-2)= log_2(3x-2)^{1/2}$ and the equation becomes $log_2(x+1)- log_2(3x-2)^{1/2}$ $= log_2\left(\frac{x+1}{(3x-2)^{1/2}}= 1$.

And that gives $\frac{x+1}{(3x-2)^{1/2}}= 2$. Now you can write that as $x+1= 2(3x-2)^{1/2}$, square both sides and finish as before.

As for $log_5 (log_x(log3 27)) = 1$ which is, indeed, the same as $log_5 (log_x(3)) = 1$, I would recommend that now you use the fact that if $log_5(A)= 1$, then $A= 5^1= 5$.

That is, you now have $log_x(3)= 5$ and that is the same as saying $3= 5^x$. Solve that for x by using whatever logarithm is handy, either "common" or "natural. $log(3)= log(5^x)= xlog(5)$ so $x=\frac{log(3)}{log(5)}$. That cannot be done "by hand". Use either common log or natural log, whichever is easier to do on your calculator.

7. Someone correct me if I'm wrong, but this is what I'm getting for that second one:

$\log_5 (\log_x (\log_3 27)) = 1$

$\log_5 (\log_x 3) = 1$

$\log_5 (\frac {\log_5 3}{\log_5 x}) = 1$

$\log_5 (\log_5 3) - \log_5 (\log_5 x) = 1$

$\log_5(\log_5 3) = 1 + \log_5 (\log_5 x)$

$\log_5(\log_5 3) = \log_5 5 + \log_5 (\log_5 x)$

$\log_5(\log_5 3) = \log_5 5(\log_5 x)$

$\log_5(\log_5 3) = \log_5 (\log_5 x^5)$

$\log_5 3 = \log_5 x^5$

$3 = x^5$

$x = \sqrt[5] {3}$