Exponential simultaneous equation with powers

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The discussion focuses on solving the exponential simultaneous equation (0.9463e^{(r-0.04)^{5/12}}) = (0.95152e^{(r-0.04)^{11/12}}) for the variable r. The solution provided indicates that r is approximately 2.90%. Participants emphasize the importance of manipulating exponents and applying natural logarithms to simplify the equation effectively.

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Hi there,Struggling with this problem, anyone know how this might be solved?

(e^(r-0.04)^(5/12))(0.9463) = (e^(r-0.04)^(11/12))(0.95152)

and solve for r

Ans: r~2.90%

Would really appreciate it!
 
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acu04348 said:
Hi there,Struggling with this problem, anyone know how this might be solved?

(e^(r-0.04)^(5/12))(0.9463) = (e^(r-0.04)^(11/12))(0.95152)

and solve for r

Ans: r~2.90%

Would really appreciate it!
It's hard to tell for certain what you're working with. Is this it?
[tex]0.9463(e^{r - .04})^{5/12} = .95152(e^{r - .04})^{11/12}[/tex]

If so, when you raise an exponent to a power, the exponents multiply, so the above can be rewritten as
[tex]0.9463e^{(5/12)(r - .04)} = .95152e^{(11/12)(r - .04)}[/tex]

Now take the natural log of both sides.
 
Thank you Mark44.
 

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