Please, can you expand this log?

Question

I'm study for a placement test, and I got this answer wrong. Can you explain me why?

log[b]((y^(2))/(xz^(5)))

log[b](y^(2)-log[b](xz^(5))

2log[b](y)-log[b](x)+log[b]z^(5)

2log[b](y)-log[b](x)+5log[b]z. This is my answer.

Old Teacher · Accepted Answer

The last plus sign should be a minus, since that term was in the denominator: 
Log [ a/(bc)] = loga -log(bc) 
= loga -[ log b+ logc ]
= log a - log b - logc 

Everything else is good.

I hope this helps!

Thomas the Mathemagician · Answer

Expanding a log--doesn't wood expand in water?  (Sorry; a real stretch there...)

Your problem is in step 2; you should have parentheses around (log_b xz^5), so step 2 should look like this: = log_b y^2 - (log_b xz^5).

Now use the Product Property on the second expression and bring down the exponent on the first one:

= 2 log_b y - (log_b x + log_b z^5) = 2 log_b y - log_b x + 5 log_b z)

...and distribute the negative sign to remove the parentheses...

==2 log_b y - log_b x - 5 log_b z.  That's where it went pear-shaped.

lia l · Answer

your  mistake  in     3rd line
 log[b](xz^(5))=log[b](x)+log[b]z^(5)

so  -log[b](xz^(5))=-log[b](x)-log[b]z^(5)

Answer would be
2log[b](y)-log[b](x) -5log[b]z

Engr. Ronald · Answer

log[b]((y^(2))/(xz^(5)))
=log[b](y^2) - [log[b](x) + log[b](z^5)]
=2log[b](y) - log[b](x) - 5log[b](z) answer//

Anonymous · Answer

you can see as the two terms are in multiplication so log will first convert it into addition .. not that you have multiplied inside

Wayne DeguMan · Answer

log(A/B) = logA - logB

so, log[b](yÂ²/xz⁵) => log[b](yÂ²) - log[b](xz⁵)

and log(AB) = logA + logB

so, log[b](yÂ²) - log[b](xz⁵) = log[b](yÂ²) - {log[b](x) + log[b]z⁵}...so, distributing the -1...

=> log[b]yÂ² - log[b]x - log[b]z⁵

Also, logA^n = nlogA

so, log[b]yÂ² - log[b]x - log[b]z⁵ => 2log[b]y - log[b]x - 5log[b]z

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Please, can you expand this log?

6 Answers