simple exponential problem

[thelannonmonk]

1. The problem statement, all variables and given/known data

5(0.10)^x = 4(0.12)^x

this problem is driving me nuts. i graphed it and got x~1.2239011 but I can't find it on paper.

i have been using the property ln(a^x) = x*ln(a)

3. The attempt at a solution

i tried doing x*ln(0.50) = x*ln(0.48)

but i still can't get it. it is making me crazy. i thought i was good at math :(

every time i try to solve it, my x's disappear. i always end up dividing ax/bx and losing them. it is driving me absolutely crazy. is there a different property i should be using? i'm almost positive that all i should need is ln(a^x) = x*ln(a)

[thelannonmonk]

also, i'm not sure if i should be getting 5x*ln(.1) = 4x*ln(.12) -or- x*ln(.5)=x*ln(.48)

[MisterX]

0.1 = e^a
ln(0.1) = a

5*(0.1)^x = 5*(e^a)^x

[thelannonmonk]

i don't think you understand my question. i have to solve for x for...

5(0.10)^x = 4(0.12)^x

when i graph 5(0.10)^x and 4(0.12)^x and use the intersection finder, i get x = 1.2239011, which when i plug into the equation yields a correct answer (.29856 = .29856)

but i simply can't find this answer on the paper.

[SteamKing]

In Sec. 3 of the OP, log (5 * 0.10^x) = log 5 + x * log (0.10) NOT log (0.5^x)
Remember, exponentiation takes precedence over simple multiplication.

[thelannonmonk]

In Sec. 3 of the OP, log (5 * 0.10^x) = log 5 + x * log (0.10) NOT log (0.5^x)
Remember, exponentiation takes precedence over simple multiplication.

http://images3.wikia.nocookie.net/__cb20091218002804/uncyclopedia/images/thumb/f/f4/Picard-facepalm.jpg/800px-Picard-facepalm.jpg

ohhh man, that's embarrassing. thanks for your help! shoulda had that one