can anyone solve this math problem (maybe with logarithms)?

Question

here is the problem:

90*(.9572^x)+
80*(.0214^x)+
80*(.0214^x) =100

now...solve for x (the equation is all one line, but the formatting gets screwed up!)

i tried to take the log of both sides, but i think that's wrong.

(note: here is why... log (a + b + c) does not equal log(a) + log(b) + log(c)... try it with a=20, b=40, c=50.... log (a + b + c) = log(110) which is not the same as log(20)+log(40)+log(50)... i think this is the key to the problem.... in the equation above, you have a similar situation where there are 3 terms that are added together....)

if anyone can show me how to solve for x it would be much appreciated!!!

Answer 1

I don't think this could be solved by logarithms. However, you can use numerical approximations to approximate the value of x...

(Do you know calculus?)

Answer 2

first combine 80*(.0214^x) + 80*(.0214^x) =
160*(.0214^x)

90*(.9572^x) + 160*(.0214^x) = 100

let z = .9572^x
let y = .0214^x

90z + 160y = 100

Answer 3

Looks like this equation can only be solved by numerical methods (Newton-Raphson), because you cannot take logarithms in this case. You have sums.
I got x= 0.661755 using the Newton's method.

Answer 4

I am not too sure about an analytical solution, but the first step is to divide both sides by 10, and then combine the two identical terms "80*...". trial and error could be a quick way, especially with a calculator or spread sheet. It may also be reasonable to rearrange the new equation, so that a^x = f(b^x), and iterate on this.

Answer 5

Remember these two relationships:

log (a^x)= x log(a)
log (a)+log(b)+log(c)=log (a*b*c)

good luck

Answer 6

3^(x-2)=9^(x+4) 3^(x-2)=3^(2(x+4)) x - 2 = 2(x + 4) x - 2 = 2x + 8 - 10 = x 3(- 10 - 2) = 9^(- 10 + 4) 3^(- 12) = 9^(- 6) 1/531441 = 1/531441