Exponential and Logarithmic Functions

BASIC CONCEPTS:

1.

Exponential functions are those functions with variable exponents. The exponent is some power over a base.

$$\begin{eqnarray*}3^x & = & 4 \\ \left (\frac{1}{2}\right)^x & = & 2y \\ y & = & 7^x \\ \end{eqnarray*}$$

2.

Simple exponential functions can always be written as a logarithmic function, where the base of the logarithm is the same as the base of the exponential function.

$$\begin{eqnarray*}3^x = 4 & \approx & \log_3{4} = x\\ \end{eqnarray*}$$

3.

Remember, the answer to any logarithmic function is always an exponent.

$$\begin{eqnarray*}\log_2{4} = 2 & because & 2^2 = 4 \\ \log_3{9} = 2 & because & 3^2 = 9 \\ \end{eqnarray*}$$

4.

The function $\ln{x}$ is the unique logarithm with base e.

$$\begin{eqnarray*}\ln{(x)} & = & \log_e{(x)} \\ \end{eqnarray*}$$

5.

Logarithmic and exponential functions are inverses IF and ONLY IF the two functions have the exact same base.

$$\begin{eqnarray*}\mbox{If}\hspace{0.5cm} f(x) = 3^x & then & f^{-1}(x) = \log_3{(x)}\\ & \downarrow &\\ \log_3{(3^x )} & = & x\\ \end{eqnarray*}$$

6.

Remember that $\ln{(x)}$ is the inverse function of e^x. Therefore,

$$\ln{(e^x )} = x$$

USING THESE CONCEPTS:



Problem 1) Solve for t:

$$\begin{eqnarray*}e^{5t} & = & 4\\ & \downarrow &\\ \ln{(e^{5t} )} & = & \ln{... ... & \ln{(4)}\\ & \downarrow &\\ t & = & \frac{\ln{(4)}}{5}\\ \end{eqnarray*}$$

Since the base of the exponential function was e, we can carefully choose a logarithm with base e to isolate the exponent. The logarithm with base e is the natural logarithm, $\ln{(x)}$.



Problem 2) Solve for x:

$$\begin{eqnarray*}5^{3x} & = & 25\\ & \downarrow &\\ \log_5{(5^{3x})} & = & \... ...w &\\ 3x & = & 2\\ & \downarrow &\\ x & = & \frac{2}{3}\\ \end{eqnarray*}$$

In this problem, since the base of the exponential function is 5, I carefully chose the logarithm with base 5. Then, when the logarithm base 5 is applied to 5^3x all that is left is 3x. This is the easiest approach. However, one could also choose the natural logarithm as well.

$$\begin{eqnarray*}5^{3x} & = & 25\\ & \downarrow &\\ \ln{(5^{3x})} & = & \ln{... ...}}\\ & \downarrow &\\ x & = & \frac{\ln{(25)}}{3\ln{(5)}}\\ \end{eqnarray*}$$

Remember that $\log_5{(25)} = \frac{\ln{(25)}}{\ln{(5)}}$ by definition. Therefore,

$$\begin{eqnarray*}x & = & \frac{\ln{(25)}}{3\ln{(5)}}\\ & \downarrow &\\ x & ... ...frac{1}{3}\right)2\\ & \downarrow &\\ x & = & \frac{2}{3}\\ \end{eqnarray*}$$

So we get the same answer with either approach. It looks clear that choosing our logarithm function to have the same base as the exponential function can be simpler for not-so-complicated equations. However, you can't go wrong with the natural logarithm, $\ln{(x)}$.

One very common mistake is the following. The function $\log_5{(7)}$ is a value. The logarithm base 2 and the 7 can not be seperated. But occasionally someone tries to simply divide by $\log_2$. The following results...

$$\begin{eqnarray*}2^{3x} & = & 4\\ & \downarrow &\\ \log_2{(3x)} & = & \log_2... ...w &\\ 3x & = & 4\\ & \downarrow &\\ x & = & \frac{4}{3}\\ \end{eqnarray*}$$

But this is incorrect since $2^4 \neq 4$.



Problem 3) Solve for y:

$$\begin{eqnarray*}\ln{(y)} & = & 5^{2x^{2} + 1}\\ & \downarrow &\\ e^{\ln{(y)... ...{2x^{2}+1}}\\ & \downarrow &\\ y & = & e^{5^{2x^{2} + 1}}\\ \end{eqnarray*}$$

This is a complicated example so look at a simpler version.....

$$\begin{eqnarray*}\ln{(y)} & = & 4\\ & \downarrow &\\ e^{\ln{(y)}} & = & e^{4}\\ & \downarrow &\\ y & = & e^{4}\\ \end{eqnarray*}$$

Because the base of the natural logarithm is e, we chose our exponential function to have base e. Then raising both sides over e allows us to isolate y for a solution. This same technique works for different base logarithms.

$$\begin{eqnarray*}\log_3{(x)} & = & 9\\ & \downarrow &\\ 3^{\log_3{(x)}} & = & 3^{9}\\ & \downarrow &\\ x & = & 3^{9}\\ \end{eqnarray*}$$

See the simlarity. Simply choose an exponential function with the same base as the logarithm and it all falls out. Try not to be intimidated by logarithms and exponential functions. Solving these equations comes down to the simple step of matching the bases of the logarithm and exponential function, especially in these simpler examples. So, if our base is e, we choose the natural logarithm to isolated the exponent. If our base is 2, we choose $\log_2$as our logarithm. Then apply that logarithm to BOTH sides of the equation and KERBLOOIE!!!! Things just start falling into place.



In the case of more complictaed word problems, it is a good idea to simply stick with the natural logarithm. So for example, if we have set up a half-life equation as such:

$$\frac{1}{3}=\left(\frac{1}{2}\right)^{t/5}$$

then it is easiest to take the natural log of both sides.....

$$\begin{eqnarray*}\ln{(\frac{1}{3})} & = & \ln{\left(\frac{1}{2}\right)^{t/5}}\\ ... ...ac{\ln{(\frac{1}{3})}}{\ln{\left(\frac{1}{2}\right)}} & = & t\\ \end{eqnarray*}$$

From here further simplification leads to the solution:

$$t=5\frac{\ln{(3)}}{\ln{(2)}}$$

CONCLUSION:


The key to working with logarithms is understanding that these functions are values whose answer is always an exponent over the base. Do not try and seperate logarithms. The term $\log_2$ makes no sense, where $\log_2(4)$ does. When working with complicated word problems, stick with the natural logarithm. But when solving simpler algebraic equalities, it may be easier to choose a logarithm with a different base equaling the base of the exponential function. Keep your bases matching and you can't go wrong.








FINAL PROBLEM:

$$\begin{eqnarray*}3^{\log_5^{x^2}} & = & 27\\ & \downarrow &\\ \log_3{(3^{\lo... ... x^{2} & = & 125\\ & \downarrow &\\ x & = & \sqrt[2]{125}\\ \end{eqnarray*}$$