Ask Me Help Desk - Properties of logarithms: solve for why in terms of x

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Properties of logarithms: solve for y in terms of x

Here is the question I have to answer:

5log(2)y – log(2)x = 3log(2)4 + log(2)a (I'm sorry if this is not written properly. I'm not sure how to do subscript typing for the 2s.)

This is what I've come up with:

log(2)y^{5} = 3log(2)2^{2} + log(2)a + log(2)x

log(2)y^{5} = 3log(2)2^{2} + log(2)ax

y = 3(2ax)^{1/5)

My textbook says the solution should be:

y = 2(2ax)^{1/5}

Where have I gone wrong?

Remember that $n log(b) = log(b^n) $

It seem that you didn't handle the $3 log_2 2^2$ properly; $3 log_2 2^2 = log_2 (2^2)^3$ .

$ 5log_2y – log_2x = 3log_24 + log_2a \\ 5log_2y = 3log_2(2^2) + log_2a + log_2x = log_2(2^2)^3 + log_2a + log_2x = log_2(2^6 ax) \\ log_2y = \frac 1 5 log_2(2^6 ax) = log_2 (2^6ax)^{1/5}\\ y = (2^6ax)^{1/5}=2(2ax)^{1/5} $

Hi. Okay, so I understand where I went wrong with the 3log(2)2^{2}... but now I am confused about the very last step. How does (2^{6}ax)^{1/5} become 2(2ax)^{1/5}?

Also, how can I type my equations like yours? I tried to type them up with Word then just copy and paste but that doesn't work.

Lynn:

$ (2^6)^{1/5} = 2^{6/5} = 2 \cdot 2^{1/5} $

To get all the neat math layout you use Latex. See: https://www.askmehelpdesk.com/math-s...las-50415.html

Aaahhh, I see! Thank you very much! Greatly appreciated!

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