Show that for any $a0$, it holds that ${{a}^{b}}={{\mathrm{e}}^{b\ln a}}$. Hence, or otherwise,

Question

Show that for any $a>0$, it holds that ${{a}^{b}}={{\mathrm{e}}^{b\ln a}}$. Hence, or otherwise, solve the equation ${{x}^{\ln 4}}+{{x}^{\ln 10}}={{x}^{\ln 25}}$.

Accepted Answer

(b) $x=\left((\sqrt5-1)/2\right)^{1/\ln(2/5)}\approx1.69$ (3 s.f.).

Exponential and Logarithmic Functions - Solving Exponential Equations Using Substitution

Exponential and Logarithmic Functions - Solving Exponential Equations Using Substitution

Solution:

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