Why is it that $\ln\Big(1+\frac{x}{y}\Big)=\ln(1+\exp(\ln x - \ln y))$?

Question

Is the relation $\ln\Big(1+\frac{x}{y}\Big)=\ln(1+\exp(\ln x - \ln y))$ an approximation?

If so, how can I derive this relation?

Answer 1

$$ \ln(x) - \ln(y) = \ln\left(\frac{x}{y} \right) \\ \exp\left(\ln\left(\frac{x}{y}\right)\right) = \frac{x}{y} \\ \implies \ln\left(1 + \frac{x}{y}\right) $$

Answer 2

It’s not an approximation; it’s exact. Since the natural logarithm and $e^x$ are inverse functions, we know that, since $f^{-1}(f(x)=x$, $$e^{\mathrm{ln} x}=x$$ Replacing $x$ with $\frac{x}{y}$, we have $$e^{\ln(\frac{x}{y})}=\frac{x}{y}$$ So, plugging this into the original equation $$\ln\left( 1 + \frac{x}{y}\right)=\ln\left(1+e^{\ln\left(\frac{x}{y}\right)}\right)$$ We know that a property of all logarithms is that $$\log_w(\frac{x}{y})=\log_w(x)-\log_w(y)$$ So, substituting we have $$\ln\left(1+e^{\ln\left(\frac{x}{y}\right)}\right)= \ln\left(1+e^{\ln(x)-\ln(y)}\right)= \ln\left(1+\exp(\ln(x)-\ln(y))\right)$$

Answer 3

Note that

$$\log a-\log b = \log\frac{a}{b}$$

Applying that here, you get

$$\ln \big(1+e^{\ln x-\ln y}\big) = \ln \big(1+e^{\ln \frac{x}{y}}\big)$$

Also, as inverses, $e^{\ln a} = a$ (and $\ln e^a = a$), meaning the expression simplifies to

$$\ln \bigg(1+\frac{x}{y}\bigg)$$

Also, to point out, this isn’t an approximation. The two expressions are exactly equivalent.