logarithm

Since \[b^{e}=x\] and \[\log_{b}(x)=e\], substituting \[e\] into our original equation yields us \[b^{\log_{b}(x)}=x\].

The definition of logarithm states that the logarithm of \[b^{x}\] to the base, \[b\], is the exponent, \[x\], which \[b\] must be raised to produce \[b^{x}\].

Let \[e=\log_{b}(x)\]. \[b^{e}=x\], thus \[(b^{e})^{r}=x^{r}\]. Applying \[\log_{b}\] to both sides, we get \[\log_{b}(b^{er})=\log_{b}(x^{r})\], which simplifies into \[\log_{b}(x^{r})=mr=\log_{b}(x)\cdot r\].

Let \[\log_{b}(x)=m,\log_{b}(y)=n\], thus \[x=b^{m},y=b^{n}\].

\begin{align*} \log_{b}(xy)&=log_{b}(b^{m}\cdot b^{n})\\ &=\log_{b}(b^{m+n})\\ &=m+n\\ &=\log_{b}(x)+\log_{b}(y)\\ \end{align*}

Let \[\log_{b}(x)=m,\log_{b}(y)=n\], thus \[x=b^{m},y=b^{n}\].

\begin{align*} \log_{b}\left( \frac{x}{y} \right)&=\log_{b}\left( \frac{b^{m}}{b^{n}} \right)\\ &=\log_{b}(b^{m-n})\\ &=m-n\\ &=\log_{b}(x)-\log_{b}(y)\\ \end{align*}

Let \[m=\log_{b}(x)\], \[b^{m}=x\]. Add \[\log_{a}\] to both sides we get \[\log_{a}(b^{m})=\log_{a}(x)\], which simplifies into \[m\log_{a}(b)=\log_{a}(x)\]. Thus \[m=\frac{\log_{a}(x)}{\log_{a}(b)}\], substituting \[m=\log_{b}(x)\] we get \[\log_{b}(x)=\frac{\log_{a}(x)}{\log_{a}(b)}\].

• derivatives of exponential and logarithmic functions
  
• derivative of natural logarithm function
  
• derivatives of general exponential and logarithmic functions