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Why $$\ln 2=\ln 1.075^t\implies \ln 2=t\ln 1.075$$

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  • $\begingroup$ It is also valid in the other direction. $\endgroup$ – Jika May 16 '14 at 16:54
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From definition we have:

$a=e^{\ln{a}}$

$a^{b}=e^{\ln(a^{b})}$

$a^{b}=(e^{\ln{a}})^{b}=e^{b\ln{a}}$ thus

$e^{\ln(a^{b})}=e^{b\ln{a}}$ $\;\iff\;$ $\ln(a^{b})=b\ln{a}$

The last follows from the injectivity of the function $e^{x}$.

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That follows from one of the laws of logarithms:

$$\ln a^b = b\ln a \tag{$*$}$$

In your case $a=1.075, \; b=t$.

$(*)$ Note that this holds for $\log a^b$, in general, not just the natural log.

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This is a specific form of logarithms of powers, or $log_b(x^r) = r\cdot log_b(x)$. To prove this statement, let $y=r\cdot log_b(x)$. Then, $$ b^y = b^{r\cdot log_b(x)} \\ b^y = (b^{log_b(x)})^r \\ b^y = x^r \\ log_b(b^y) = log_b(x^r) \\ y = log_b(x^r) $$

Therefore, the original statement is proved. Your specific example follows for $x=1.075$, $b=e$, and $r=t$ since setting both equations equal to $ln(2)$ reduces to $ln(1.075^t) = t\cdot ln(1.075)$.

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