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christinaxxx
 3 years ago
how do you solve Log5 (2x) – 1 = log5 (2x7)
christinaxxx
 3 years ago
how do you solve Log5 (2x) – 1 = log5 (2x7)

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jennychan12
 3 years ago
Best ResponseYou've already chosen the best response.0is that log base 5? if so just divide out the bases so you get 2x = 2x7 solve for x

christinaxxx
 3 years ago
Best ResponseYou've already chosen the best response.0it is base five, but what about the minus 1?

jennychan12
 3 years ago
Best ResponseYou've already chosen the best response.0move the logbase5(2x7) to the other side and move the 1 over. try from there?

christinaxxx
 3 years ago
Best ResponseYou've already chosen the best response.0i did and i got x=3 and i feel like that's wrong

jennychan12
 3 years ago
Best ResponseYou've already chosen the best response.0really? i got x = 3.

christinaxxx
 3 years ago
Best ResponseYou've already chosen the best response.0this is what i did. 1=logbase5 (2x+7/2x) (2x+7)/(2x)= 5^1 (2x+7)/(2x)=1/5 then i cross multiplied and got 5(2x+7)=1(2x) 10x+35=2x 11x+35=2 11x=33 x=3 righttttt?

jennychan12
 3 years ago
Best ResponseYou've already chosen the best response.0Log5 (2x) – 1 = log5 (2x7) i meant like move the 1 to the right and the logbase5(2x7) to the left so you get (2x)/(2x7) = 5 i got 3, but neither 3 nor 3 work as answers. and your 2x+7 should be a 7...unless u wrote the question wrong...

christinaxxx
 3 years ago
Best ResponseYou've already chosen the best response.0nope, my question is from a worksheet. no error in my copying it down. but the negative one would be the exponent of the base 5, wouldn't it? so you'd have to raise five to the negative one?

jennychan12
 3 years ago
Best ResponseYou've already chosen the best response.0i dunno why we're getting diff answers cuz u could do it multiple ways
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