Solving Exponential Functions Help

Subject Areas

Tools / Misc

bCalc

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scrnGC3: f(x) Plot Input

scrnGC4: g(x) Plot Input

The inputs above will be hidden when the program finally gets launched. Currently, I am not hiding them even when you close the calculator. The purpose of the inputs is to help in debugging. This allows you to enter simple code on the calculator screen but the program translates the input to javascript math. Give it a try and see if you notice any obvious problems or have suggestions for improvements.

gCalc

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Select precision* of answer:

Select angle mode:

Enter expression to evaluate here:

The answer is shown below:

Enter function to be plotted:

f(x) =

Enter 2nd function, if needed:

g(x) =

Change or reset the x-y scaling below:

x_min =

x_max =

y_min =

y_max =

Accuracy Disclaimer

The javascript math object methods used for this calculator are not accurate for very large or very small numbers. In particular, the trigonometric functions may yield very small answers that should actually be zero or very large answers that should be infinite. This typically occurs at 90� or pi/2 intervals.

DO NOT USE WHERE HIGH PRECISION IS REQUIRED. USE AT YOUR OWN RISK.

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Solving Exponential Functions Help

Solving exponential functions can be greatly simplified if you know how to use logarithms. This lesson will begin by using exponents and/or radicals since it is quite common to have to solve these functions prior to learning about logarithms. The lesson will be concluded using logarithms, at which point, you can decide for yourself, which is easier.

The trick to solving exponential functions is to ensure that you a common base. If you have an equation such as 3 ⁴ = 3 ^x, it is relatively easy to see that since the bases are the same, the exponents must be the same, so x = 4. Many times the given bases are not the same but with some factoring and simplification a common base can be found.

Example: Given: 8 ² = 4 ^{x + 1}

( 2 ³ ) ² = ( 2 ² ) ^{x + 1}

2 ⁶ = 2 ^{2x + 2}

6 = 2x + 2

2x = 4

x = 2

Although the original problem had two different bases, 8 and 4, it was not difficult to find the common factor, 2. Once both sides of the equation had a common base solving for x was rather straight forward.

But what do you do if it is not possible to find a common base, such as in the following problem. This is where logarithms come to the rescue.

Example: Given: 5 ² = 3 ^{x + 1} ( 5 ² = 25 and 3³ = 27, so x is close to 2 )

2 log (5) = ( x + 1 ) log ( 3 )

x + 1 = 2 log (5) / log ( 3 ) = 2.93

x = 1.93

No single method leads to the quickest solution every time, which is why you need to gain proficiency using both methods. Simple problems can usually be done quickest using exponents but difficult problems generally require the use of logarithms.