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Quick math question

So I'm solving logarithmic equations. How do I simplify an expression when there are two logarithms that share the same number that they're being raised to, while their bases are powers of each other?

I realize the way I explained that probably sounded confusing. As a simple example, how would I simplify/combine this into one logarithm?

log_7(x)+log_49(x)
log base 7 to get x+log base 49 to get x

Any help would be greatly appreciated.

January 16, 2014

4 Comments • Newest first

kcwt

[quote=iDrinkOJ]how do you get the (2+1) part?[/quote]

When he changed base the expression became log_49(x) + 2log_49(x) --> Since you were dividing by half which is the same as multiplying by 2
Then he just took out log_49(x) as a common factor to get log_49(x) * (2 + 1) = 3 * log_49(x)
He could have just added them together though....

Reply January 17, 2014 - edited
iDrinkOJ

[quote=SouIStrike]log_7(x) = log_49(x) / log_49(7) [change of base]

log_49(7) = 1/2 [since sqrt(49) = 7]

then you can factor: log_49(x) * (2 + 1) = 3 * log_49(x)[/quote]

how do you get the (2+1) part?

Reply January 17, 2014 - edited
TheDarkHero3

log_49(x)=(log_7(x))/2 If you think about it, 49 can be expressed as 7^2, so since 7 is the square root of 49, we divide log_7 by 2.
log_7(x)+log_49(x)=log_7(x)+.5*log_7(x)=1.5*log_7(x)
EDIT: Nvm, you wanted the answer in base 49, so you just turn base 7 into 2*base 49, like the poster above did.

Reply January 16, 2014 - edited
SouIStrike

log_7(x) = log_49(x) / log_49(7) [change of base]

log_49(7) = 1/2 [since sqrt(49) = 7]

then you can factor: log_49(x) * (2 + 1) = 3 * log_49(x)

Reply January 16, 2014 - edited