Exponential Equations

[mathxyz]

I need to solve for x for BOTH questions. I don't know where to start or how to tackle this question.

1)....(3/5)^(x) = 7^(1 - x)

2)....1.2^(x) = (0.5)^(-x)

 

[tkhunny]

1)....(3/5)^(x) = 7^(1 - x)

Are you allowed to use logarithms?

(3/5)^(x) = 7^(1 - x)
log((3/5)^(x)) = log(7^(1 - x))
x*log(3/5) = (1-x)*log(7)
x*log(3/5) = log(7) - x*log(7)
x*log(3/5) + x*log(7) = log(7)
x*(log(3/5) + log(7)) = log(7)
x = log(7)/(log(3/5) + log(7))

It isn't pretty.

You do the other one.

 

[mathxyz]

I get it...

I have to take the log of both sides, right?

 

[ChaoticLlama]

Re: I get it...

I have to take the log of both sides, right?

Yes, you take the log of both sides.

 

[mathxyz]

ok

Thanks

 

[Denis]

1)....(3/5)^(x) = 7^(1 - x)

Another way is rearrange the terms first, and end up with the "log" job:
I'll assume you know stuff like 2^(-3) = 1 / 2^3

3^x / 5^x = 7[7^(-x)]

3^x / [5^x * 7^(-x)] = 7

3^x * 7^x / 5^x = 7

21^x / 5^x = 7

(21/5)^x = 7

x = log(7) / log(21/5)

 

[mathxyz]

Thanks

Thank you all for your great tips and notes.