Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

I need to simply simplify: log100^4x Can you help

OpenStudy (anonymous):

Would it be 8x?

OpenStudy (phi):

what base is the log?

OpenStudy (anonymous):

I'm assuming 100...all I'm given is log100^4x

OpenStudy (phi):

I would assume 10. so the question would be \[ log_{10}(100^{4x} )\]

OpenStudy (phi):

there is this log property: log(a^b)= b*log(a) I would use that first.

OpenStudy (anonymous):

so 4x?

OpenStudy (phi):

Do it in steps. Can you use log(a^b)= b*log(a) (match to your problem) to rewrite log(100^(4x))

OpenStudy (anonymous):

log100=2

OpenStudy (anonymous):

then 2^4x

OpenStudy (phi):

If the base is 10, that is true. In this case, it is a safe bet they mean log base 10

OpenStudy (phi):

But back to the question: Can you use log(a^b)= b*log(a) (match to your problem) to rewrite log(100^(4x)) If you understand how to do this, you will be able to solve all these type problems. To use the rule log(a^b) =b*log(a) match a and b to your problem.

OpenStudy (phi):

btw then 2^4x is not the correct answer.

OpenStudy (anonymous):

log(100^(4x)) would = 4xlog100.....correct?

OpenStudy (phi):

Yes. Definitely on the right track. now we can simplify \[ log_{10}(100) =\]

OpenStudy (anonymous):

2

OpenStudy (phi):

so put it all together: 4*x*2=

OpenStudy (anonymous):

8x

OpenStudy (phi):

That's how you do this type problem. Hope it makes sense.

OpenStudy (anonymous):

So I'm correct with 8x?

OpenStudy (phi):

yes log(100^(4x)) = 4x log(100)= 4x*2 = 8x

OpenStudy (anonymous):

sweet!! Thank you!! I appreciate your time!

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!