Rearrange to find sigma

It's supposed to be to the power of -(1 / o), can't seem to show it on Latex for some reason. Anyway, the question asks to isolate o. My working:

$C_1^{-\frac{1}{o}} = B(1 + r_1) C_2^{-\frac{1}{o}}$

$\displaystyle \left(\frac{C_2}{C_1}\right)^{ \frac{1}{o} } = B(1 + r_1)$

$\dfrac{C_2}{C_1} = B^o (1 + r_1)^o$

Then, the solution seems to say that I can eliminate $C_2$ by setting it equal to $(1 + G)(C_1)$

Which gets me:

$\dfrac{(1+G)C_1}{C_1} = 1 + G = B^o (1+r_1)^o$

Now, for the confusing part. Firstly, the lecturer seemed to use the fact that $\ln(1 + x) = x$ (testing this in my calculator disproves this for some reason). And secondly, I have no idea what happened to the 1 in the middle side.

$G = o\ln B + o r_1$

Any help would be appreciated

Firstly: if you want to get the exponent formatted properly on LaTeX, use left and right braces { } instead of brackets ( ).


Your lecturer uses the approximation that $\ln (1+x) \sim x$ for very small values of $x$ (try it on your calculator). This approximation comes from truncating the Maclaurin series for $\ln (1+x) = x + \mathcal{O}(x^2)$. I'm not an economist and have no understanding of the actual economics happening here but presumably $G$ and $r_1$ are sufficiently small here for the approximation to be valid.


Your lecturer now takes the equation $1 + G = B^o (1+r_1)^o$ and takes logarithms of both sides to get $\ln (1+G) = o\ln B + o \ln (1+r_1)$ and uses the approximation $\ln (1+x) \sim x$ for the terms with the logarithms.