Question 277239: Solve the equation exactly for x.
e^x+3=8*6^x
Answer by jsmallt9(3759)   (Show Source):
You can put this solution on YOUR website! I'm going to assume that the equation is
because I don't see how to solve  . If I am correct, then in the future please put parentheses around multiple term exponents, numerators, denominators and arguments to functions.
With variables in exponents we usually use logarithms to solve the equation. The base of the logarithm can be anything but it is often to your advantage to choose a base for the logarithm that matches the base for the exponent with the variable in it.
Your equation has two exponents with a variable in them so we have a choice of base e (ln) or base 6 logarithms. The problem says to solve for the exact value of x. This is code for "Don't use your calculator" because calculators provide only decimal approximations for most logarithms. So there is little advantage to one base over the other. (If decimal approximations were desired, then the base e logarithm would be the best choice because many calculators "do" base e (ln) logarithms and no calculators do base 6 logarithms.) I'll do the problem using both logarithms so you can see both ways.
Find the base e logarithm of each side:
On the left side we can use the property of logarithms,  , to move the exponent in the argument out in front. On the right side we cannot do the same thing (yet) because the exponent in the argument applies just to the 6, not to the entire argument.
On the right side we need to separate the 8 and the  in the argument. To our rescue comes another property of logarithms:  . Using this on the right side we get:
Now we can use the other property to move the exponent out in front:
(x+3)ln(e) = ln(8) + x*ln(6)
(Note: The property that we just used to move the exponent out in front is exactly the reason we use logarithms on problems with variable in the exponents! It allows us to move the variables out of the exponent.)
The ln(e) is 1 by definition so the equation simplifies to:
x+3 = ln(8) + x*ln(6)
Now that the x's are out of exponents, all that is left is to solve for x. For this we need to gather the terms with x on one side and the terms without x on the other. Subtracting x and subtracting ln(8) from each side we get:
3 - ln(8) = x*ln(6) - x
On the side with the x terms we need to factor out x:
3 - ln(8) = x*(ln(6) - 1)
And finally divide both sides by (ln(6) - 1):
which is an exact solution for x. (Note: You cannot cancel the 3 in ln(6) with the 3 in the numerator! Nor can you cancel the 2 in ln(6) with the 2 in ln(8)! This fraction is a simple as it can get.)
Solving the problem with base 6 logarithms instead of base e (without the commentary):
Although this looks very different from our base e solution, it is just as exact, just as correct.
|
|
|
