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by **crescentcitycid** on Sun Feb 27, 2011 9:32 pm

I have a series of problems that state: given that log_b(2)=.39, log_b(3)=.61 and log_b(5)=.9, find the following for logs, do no try to find "b"

Then there are easy ones to solve using multiplication and division properties, but there is one I'm not sure about.

b^log_b(5)

Can that go any further than b^.9 ?

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by **stapel_eliz** on Sun Feb 27, 2011 10:10 pm

crescentcitycid wrote:...there is one I'm not sure about.

b^log_b(5)

Can that go any further than b^.9 ?

Yes. In fact, avoid that entirely. Instead of simplifying the exponent on $b$ , try setting up and solving an exponential equation:

. . . . . $b^{\log_b(5)}\, =\, x$

...for some numerical value $x.$ Then:

. . . . . $\log_b\left(b^{\log_b(5)}\right)\, =\, \log_b(x)$

Apply a log rule to get:

. . . . . $\log_b(5) \times \log_b(b)\, =\,...?$

Simplify the left-hand side, and the answer should pop out at you. :wink:

by **crescentcitycid** on Sun Feb 27, 2011 10:18 pm

.9
Awesome, thank you.

by **stapel_eliz** on Sun Feb 27, 2011 10:37 pm

crescentcitycid wrote:.9

No. Try simplifying the left-hand side. Then look at what you have left.

by **crescentcitycid** on Sun Feb 27, 2011 10:51 pm

Well if log_b(5)=.9 and log_bb=1 then .9(1)=.9

by **stapel_eliz** on Mon Feb 28, 2011 12:48 am

Start with the equation I gave you earlier. Follow the instructions.

What is the expression for the right-hand side? What is the simplified (not "evaluated") form of the left-hand side?

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