# If f(x)=\log_ax, show that \frac{f(x+h)-f(x)}{h}=\log_a\left(1+\frac{h}

If $f\left(x\right)={\mathrm{log}}_{a}x$, show that $\frac{f\left(x+h\right)-f\left(x\right)}{h}={\mathrm{log}}_{a}{\left(1+\frac{h}{x}\right)}^{\frac{1}{h}},h\ne c0$
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Arham Warner
Step 1
The difference quotient $\frac{f\left(x+h\right)-f\left(x\right)}{h}$ can be used to calculate the first derivative of the function by taking the limit $h\to 0$. It will give f'(x) if it exists.
For the given problem, two properties of logarithms will be used. The first is the difference property, which is ${\mathrm{log}}_{a}x-{\mathrm{log}}_{a}y={\mathrm{log}}_{a}\frac{x}{y}$. The second one is the exponent property of logarithms, which is $y{\mathrm{log}}_{a}x={\mathrm{log}}_{a}{x}^{y}$
Step 2
Substitute $f\left(x\right)={\mathrm{log}}_{a}x$ in the difference quotient $\frac{f\left(x+h\right)-f\left(x\right)}{h}$ and simplify using the properties of logarithms.
$\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{{\mathrm{log}}_{a}\left(x+h\right)-{\mathrm{log}}_{a}x}{h}=$
$=\frac{{\mathrm{log}}_{a}\frac{x+h}{x}}{h}$
$=\frac{1}{h}{\mathrm{log}}_{a}\left(1+\frac{h}{x}\right)$
$={\mathrm{log}}_{a}{\left(1+\frac{h}{x}\right)}^{\frac{1}{h}}$
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