1. ## Logarithms help

Well I have 2 questions but they are labeled (i) and (ii) from the same question so I'll post them in a single thread. Please forgive me if I'm not supposed to that because I just joined today and I'm not too familar with this forum.
I don't understand how to type in latex so I will attach pictures.

These are the steps I used for part (i):

And for part (ii):

The answer key says x = 2 which I do not quite understand because there are two part questions, so I assume it is wrong. My exams are coming soon and I find that I often come across this kind of problem so any help would be much appreciated, thanks!

2. Originally Posted by caramelcake
Well I have 2 questions but they are labeled (i) and (ii) from the same question so I'll post them in a single thread. Please forgive me if I'm not supposed to that because I just joined today and I'm not too familar with this forum.
I don't understand how to type in latex so I will attach pictures.

These are the steps I used for part (i):

And for part (ii):

The answer key says x = 2 which I do not quite understand because there are two part questions, so I assume it is wrong. My exams are coming soon and I find that I often come across this kind of problem so any help would be much appreciated, thanks!
$\displaystyle \log_2(x) + \log_8(2x) = \frac{7}{3}$

$\displaystyle \log_8(2x) = \frac{\log_2(2x)}{\log_2(8)} = \frac{\log_2(2)+\log_2(x)}{3} = \frac{1+\log_2(x)}{3}$

$\displaystyle \log_2(x) + \frac{1+\log_2(x)}{3} = \frac{7}{3}$

Multiply by 3

$\displaystyle 3\log_2(x)+1+\log_2(x) = 7$

$\displaystyle \log_2(x^3) + \log_2(x) = 6$

$\displaystyle \log_2(x^4) = 6$

3. Thank you e^(i*pi) for your reply, but I do believe that your reply was exactly what I had already come to in my first post. Or is it that I can simply put $\displaystyle \log_2(x^4) = 6$ as my answer?

4. $\displaystyle \log_2{(x^4)} = 6$

$\displaystyle x^4 = 2^6$

$\displaystyle x = (2^6)^{\frac{1}{4}}$

$\displaystyle x = 2^{\frac{3}{2}}$

$\displaystyle x = 2\sqrt{2}$.

5. Thank you for your kind help! I didn't know that you could 'transfer' exponents like that, maybe that's why I'm always stuck when solving logarithm problems. Thanks again!

6. $\displaystyle \log{(4^x - 4)} - x\log{2} = \log{3}$

$\displaystyle \log{(4^x - 4)} - \log{(2^x)} = \log{3}$

$\displaystyle \log{\left(\frac{4^x - 4}{2^x}\right)} = \log{3}$

$\displaystyle \frac{4^x - 4}{2^x} = 3$

$\displaystyle \frac{(2^2)^{x} - 2^2}{2^x} = 3$

$\displaystyle 2^{2x} - 2^2 = 3(2^x)$

$\displaystyle (2^x)^2 - 3(2^x) - 4 = 0$

$\displaystyle (2^x - 4)(2^x + 1) = 0$

$\displaystyle 2^x - 4 = 0$ or $\displaystyle 2^x + 1 = 0$.

But since $\displaystyle 2^x > 0$ for all $\displaystyle x$, that means the only case that works is

$\displaystyle 2^x - 4 = 0$

$\displaystyle 2^x - 2^2 = 0$

$\displaystyle 2^x = 2^2$

$\displaystyle x = 2$.

7. ## Cased closed!

Thank you e^(i*pi) and Prove It for your quick help, I really appreciate it!