Ask your own question, for FREE!
Mathematics 19 Online
OpenStudy (abbles):

Which of the following expressions is equivalent to the one shown below? (check my answer)

OpenStudy (abbles):

\[2\log_225 - 2\log_25 + \log_23\]

OpenStudy (abbles):

Answer choices: \[\log_275\] \[\log_243\] \[2\log_215\] \[2\log_223\]

OpenStudy (abbles):

I chose answer C.

OpenStudy (abbles):

@sweetburger long time no talk! do you have a minute for an old friend? :D

OpenStudy (bootyanonymous):

the first choice should be the right one

OpenStudy (abbles):

Yeahh it was :/ can you explain it please?

OpenStudy (abbles):

Nice name btw xD

OpenStudy (bootyanonymous):

\[(\frac{ 2\log(25) }{ \log(2) }) - (\frac{ 2\log(5) }{ \log(2) }) + (\frac{ \log(3) }{ \log(2) })\]

OpenStudy (bootyanonymous):

i didn't use a calculator but if you want to make your life easier then use one

OpenStudy (bootyanonymous):

oh and thx for the compliment about my username :D

OpenStudy (abbles):

And then what? is there a way to calculate it without a calculator?

OpenStudy (sweetburger):

Ok there is a simpler way to do this I think.

OpenStudy (sweetburger):

First write it as \[\log_{2}(25^2)-\log_{2}(5^2)+\log_{2}(3) \] now because each term has the same base we can combine them. \[\log_{2}(\frac{ 225 }{ 25 })+\log_{2}(3) \] this then reduces to \[\log_{2}(25 )+\log_{2}(3) \] which then can be combined to form \[\log_{2}(75) \]

OpenStudy (sweetburger):

Typo* for the second step it should be \[\log_{2}(\frac{ 625 }{ 25 } )+\log_{2}(3) \]

OpenStudy (sweetburger):

@Abbles sorry for not responding sooner I was afk.

OpenStudy (bootyanonymous):

to be honest i totally forgot about the way you did it :/

OpenStudy (sweetburger):

I mean the change of base formula is a valid way to do it except it will be hard to come to an exact answer. You would probably be stuck with a rational approximation.

OpenStudy (bootyanonymous):

true

OpenStudy (abbles):

@sweetburger thank you so much! that was extremely helpful. :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!