Solving a Logarithm Problem with Given Logs: Step-by-Step Guide

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The discussion focuses on solving the logarithmic expression log[a]((ax²)/(√y)) - 2, given log[a]x = 5 and log[a]y = 8. The solution involves applying logarithmic properties such as log_a(p^q) = q log_a(p), log_a(pq) = log_a(p) + log_a(q), and log_a(p/q) = log_a(p) - log_a(q). The final result is -14, derived through a step-by-step application of these logarithmic rules.

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If log[a]x=5 and log[a]y=8, solve:

log[a]((ax2)/(√y))-2

---------

I am completely lost. I've tried some ways of doing this question but I can't get past the second and third steps. This is one of the last questions in my homework and I do not have a step-by-step solutions manual, only the final answer which would be useless because I will have no idea how to get there. Can someone please give me a step-by-step solution? Please and thanks!
 
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Alaba27 said:
If log[a]x=5 and log[a]y=8, solve:

log[a]((ax2)/(√y))-2

---------

I am completely lost. I've tried some ways of doing this question but I can't get past the second and third steps. This is one of the last questions in my homework and I do not have a step-by-step solutions manual, only the final answer which would be useless because I will have no idea how to get there. Can someone please give me a step-by-step solution? Please and thanks!

Welcome to MHB, Alaba27! :)

There are a couple of calculation rules for logarithms.

In particular:
$$\log_a p^q = q \log_a p \\
\log_a pq = \log_a p + \log_a q \\
\log_a \frac p q = \log_a p - \log_a q \\
\sqrt{p} = p^{1/2}$$
Can you apply those?
 
You might need to use :

$$\log_a a = 1$$
 
I just don't understand how to use those formulas with this kind of the question. None of the other questions in my homework are in that format and it's extremely confusing. This is what it looks like.

View attachment 730
 

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I got the solution! After multiple attempts and help from others I got this:

= (ax2/y1/2)-2
= (a-2[x2]-2)/([y1/2]-2
= (a-2x-4)/(y-1)
= y/a2x4

loga(y/a2x4) = -2[loga(a) + 2loga(x) – 1/2loga(y)]

= -2 -4loga(x) + loga(y)
= -2 – 4(5) + 8
= -2 – 20 + 8
= -14
 
Yes, good work! (Yes)

For the benefit of other students who may read this topic, I will write out a solution method using $\LaTeX$:

If $$\log_a(x)=5$$ and $$\log_a(y)=8$$, find the value of $$\log_a\left(\left(\frac{ax^2}{\sqrt{y}} \right)^{-2} \right)$$.

$$\log_a\left(\left(\frac{ax^2}{\sqrt{y}} \right)^{-2} \right)=-2\log_a\left(\frac{ax^2}{\sqrt{y}} \right)=$$

$$-2\left(\log_a(ax^2)-\log_a(\sqrt{y}) \right)=-2\left(\log_a(a)+\log_a(x^2)-\log_a(y^{\frac{1}{2}}) \right)=$$

$$-2\left(1+2\log_a(x)-\frac{1}{2}\log_a(y) \right)=-2\left(1+2\cdot5-\frac{1}{2}\cdot8 \right)=-2\left(1+10-4 \right)=-2(7)=-14$$
 

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