Moved - Help! Solve the following logarithm with steps?

[TheShockgloss]

If log₁₀ X = a and log₁₀ Y = b, Show log₁₀ (X³/Y³) using a and b?

 

[DrPhil]

If log₁₀ X = a and log₁₀ Y = b, Show log₁₀ (X³/Y³) using a and b?

Where are you stuck? We need to see your work.

What rules do you know for using logarithms? For instance, if numbers are multiplied? divided? taken to powers? If you don't have those properties memorized, re-read that chapter! Those will be the "steps" in the evaluation.

 

[wjm11]

If log₁₀ X = a and log₁₀ Y = b, Show log₁₀ (X³/Y³) using a and b?

This is not calculus. This just requires using basic log rules.

I'll get you started: log(X^3/Y^3) = log (X/Y)^3 = 3(log (X/Y))

Your turn. Please show your work.

 

[TheShockgloss]

If log₁₀ X = a and log₁₀ Y = b, Show log₁₀ (X³/Y³) using a and b?

log₁₀ X = a ; X = 10^a
log₁₀ Y = b ; Y = 10^b

Therefore;
= log₁₀ (10^3a/10^3b)
= 3a lg 10 - 3b lg 10
{lg 10 = 1}
= 3a - 3b
= 3(a-b)

 

[wjm11]

log₁₀ X = a ; X = 10^a
log₁₀ Y = b ; Y = 10^b

Therefore;
= log₁₀ (10^3a/10^3b)
= 3a lg 10 - 3b lg 10
{lg 10 = 1}
= 3a - 3b
= 3(a-b)

Correct.
Here is a different (easier?) approach, starting from where I left off:

3(log(X/Y))
=3(logX - logY)
=3(a - b)

 

[JeffM]

If log₁₀ X = a and log₁₀ Y = b, Show log₁₀ (X³/Y³) using a and b?

You have really made this problem more complex than it needs to be.

og₁₀ X = a ; X = 10^a Yes but so what?
log₁₀ Y = b ; Y = 10^b Yes but so what?

Therefore;
= log₁₀ (10^3a/10^3b) This step is correct but unnecessary
= 3a lg 10 - 3b lg 10
{lg 10 = 1} This step is correct but unnecessary
= 3a - 3b
= 3(a-b)

Use the laws of logarithms

\(\displaystyle log_{10}\left(\dfrac{X^3}{Y^3}\right) = log_{10}(X^3) - log_{10}(Y^3) = 3log_{10}(X) - 3log_{10}(Y) = 3a - 3b = 3(a - b).\)