Solving for $\theta$ using log

Question

0.88 = 0.95 x 0.5 x ($e ^{-0.06129 + \theta} + e^ {-0.04129 + \theta}$)

Does this equate to
log($\frac {0.88 \times 2}{0.95}$) = log($e ^{-0.06129 + \theta}) + log( e^ {-0.04129 + \theta}$)

Answer 1

We can be a little clever here and use the fact that $e^{a+b}=e^ae^b$. Rewriting your equation yields: $$0.88=0.95\cdot0.5\cdot\left(e^\theta e^{-0.06129}+e^\theta e^{-0.04129}\right)$$ From here, you can factor out the $e^\theta$ and solve for $\theta$.

Answer 2

Is $\log x + \log y = \log (x+y)? $

No..

$\log x + \log y = \log (x y). $

A numerical solution that finds a root is required.

Answer 3

Your starting equation is $$0.88 = 0.95*0.5 * (e ^{-0.06129 + \theta} + e^ {-0.04129 + \theta})$$ which then equals $$\frac{0.88*2}{0.95} = e ^{-0.06129 + \theta} + e^ {-0.04129 + \theta}$$.

This is the farthest we can go because $log(a+b)$ is not equal to $log(a) + log(b)$. For example, if we have $log(10+10) = log(20)$, this is not equal to $log(10) + log(10) = 1 + 1 = 2$.

We need to rewrite your solution from here on. We can factor out $e^\theta $ from this equation as doobdood noted $$\frac{0.88*2}{0.95} = e ^{-0.06129 + \theta} + e^ {-0.04129 + \theta}$$.

to get $$\frac{0.88*2}{0.95} = e^\theta*( e ^{-0.06129} + e^ {-0.04129 })$$. Now we can simplify this further to get $$\frac{0.88*2}{0.95*( e ^{-0.06129} + e^ {-0.04129 })} = e^\theta$$. Now take the natural logarithm of both sides to get $\theta = \operatorname{ln}(\frac{0.88*2}{0.95*( e ^{-0.06129} + e^ {-0.04129 })})$

$\operatorname{ln}(x)$ is the logarithm with base e, aka the natural logarithm, by the way