Sometimes a variable may be in a power. In such cases, the equation is called [color=#0000ff]exponential equation[/color]. We need [color=#0000ff]logarithms[/color] to take the unknown variable down from the power.[br][br][br]   [math]\LARGE \textcolor{blue}{y=a^x\;\;\Leftrightarrow\;\;x=\log_a(y)}[/math].[br][br]From the definition, we can easily see some basic properties of logarithmic:[br][br][br]   [math]\LARGE\begin{eqnarray}[br]a^1=a &\Leftrightarrow&\log_a(a)=1\\[br]a^0=1 &\Leftrightarrow&\log_a(1)=0[br]\end{eqnarray}[/math][br][br][br]Logarithms are sometimes thought to be difficult concept. In fact, they are very easy, if you read them right. For example, [math]\Large\textcolor{blue}{\log_a(x)}[/math] is read: [color=#0000ff]to which power should the [i]a[/i] be raised to get [i]x[/i][/color].[br][br][br]Commonly used symbol for logarithmic of base [i]e[/i] is [color=#0000ff]ln[/color] and for logarithmic of base 10 is [color=#0000ff]lg[/color] . These symbols are used during the lessons. Check your own calculator, how they are denoted there. [br][br][br]

  [math]\Large\begin{eqnarray}[br]\log_2 8=3&\text{as}&2^3=8\\[br]\lg 100=2&\text{as}&10^2=100[br]\end{eqnarray}[/math][br]

[br]  [math]\Large\begin{eqnarray}[br]\log_4 (x-2)=3 &\Leftrightarrow&x-2=4^3 &\Leftrightarrow& x=64+2=66\\[br]\log_x (81)=2&\Leftrightarrow& x^2=81&\Leftrightarrow& x=9[br]\end{eqnarray}[/math][br][br][br]

There are only three basic formulas for handling logarithms and one transformation formula for changing the base.[br][br] [b][size=150] [color=#0000ff]If [i]x[/i] > 0, [i]y[/i] > 0 and [i]r[/i] is a real number, then [/color][/size][/b][br][br] [br]  [math]\LARGE\begin{eqnarray}[br]\textcolor{blue}{1.}&&\textcolor{blue}{\log_a(xy)}&\textcolor{blue}{=}&\textcolor{blue}{\log_a(x)+\log_a(y)}\\[br]\textcolor{blue}{2.}&&\textcolor{blue}{\log_a(\frac{x}{y})}&\textcolor{blue}{=}&\textcolor{blue}{\log_a(x)-\log_a(y)}\\[br]\textcolor{blue}{3.}&&\textcolor{blue}{\log_a(x^r)}&\textcolor{blue}{=}&\textcolor{blue}{r\log_a(x)}\\[br]\end{eqnarray}[/math][br][br] [br][br]You can read the formulas also from the right to the left.[br][br][br]The base of a logarithm can be changed with the formula[br][br]   [math]\LARGE \textcolor{blue}{\log_a(x)=\frac{\log_b(x)}{\log_b(a)},}[/math][br][br]where [i]a[/i] is the original base and [i]b[/i] is the new base.[br][br]

Let us solve [math]\Large e^{x}=4.[/math][br][br]Because both sides are positive , we can take the natural logarithm. By [br]doing it, we get the variable down from the exponent and it can be [br]solved.[br][br][math]\Large\begin{eqnarray}[br] &e^{x} &=4\;\;\;(>0)\\[br]\Leftrightarrow & \ln(e^{x}) &=\ln(4)\\[br]\Leftrightarrow &x\underbrace{\ln(e)}_{=1} &=\ln(4)\\[br]\Leftrightarrow & x &=\ln(4)\\[br] \end{eqnarray}[/math][br]

Let us solve [math]\Large 3^{x}=27.[/math][br][br]Because both sides are positive , we can take the natural logarithm. By [br]doing it, we get the variable down from the exponent and it can be [br]solved.[br][br]  [math]\Large\begin{eqnarray} [br]\;& 3^{x} &=27\\[br]\Leftrightarrow & \ln(3^x) &=\ln(27)\\[br]\Leftrightarrow &x\ln(3) &=\ln(27)\\[br]\Leftrightarrow & x &=\frac{\ln(27)}{\ln(3)}\\[br]\Leftrightarrow & x &=3&|\text{Notice: } 3^3=3\cdot3\cdot3=27\\[br]\end{eqnarray}[/math][br]

Let us solve [math]\Large 5.6^{x+1}=1500.[/math][br][br]Because both sides are positive , we can take the natural logarithm. By [br]doing it, we get the variable down from the exponent and it can be [br]solved.[br][br]  [math]\Large\begin{eqnarray}[br] &5.6^{x+1} &=1500\\[br]\Leftrightarrow & \ln(5.6^{x+1}) &=\ln(1500)\\[br]\Leftrightarrow &(x+1)\ln(5.6) &=\ln(1500)\\[br]\Leftrightarrow & x+1 &=\frac{\ln(1500)}{\ln(5.6)}\\[br]\Leftrightarrow & x &=\frac{\ln(1500)}{\ln(5.6)}-1\\[br]\Leftrightarrow &x&\approx 3.2[br]\end{eqnarray}[/math][br]