Solving for x in Simple Exponential Problem

[thelannonmonk]

5(0.10)^x = 4(0.12)^x... x = ? getting aggravated

Homework Statement

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)

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.

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