You can put this solution on YOUR website! I assume the equation is
Please put exponents in parentheses. It makes the problem clear and tutors are more likely to help if the problem is clear.
Solving equations where the variable is in one or more exponents usually involves use of logarithms. We can use any base for the logarithm we use. But if we choose base 5 or base 3, we will end up with a simpler expression for the answer. And if we choose a base our calculator "knows" (like base 10 or base e (aka ln)), then it will be easier to find a decimal approximation of the answer. I will do the problem both ways so you can see the difference.
Using base 5 logarithms.
Find the base 5 logarithm of each side:
Now we can use a property of logarithms, , to "move" the exponents in the arguments out in front of the logarithms. (It is this property of logarithms which is the reason we use logarithms on problems like this. The property allows us to move the exponents out in front where we can then use regular Algebra to solve for the variable.)
By definition, so our equation becomes:
Next we will gather the x terms on one side (by subtracting x from each side):
Then we factor out x on the right side:
And finally we divide both sides by :
which is an exact expression for the solution to your equation.
Using base e (ln) logarithms.
(For the most part the steps will be similar to the ones we used above.)
Find the base e logarithm of each side:
Move the exponents in front:
(x-1)*ln(5) = 2x*ln(3)
This time, however, one of the logs does not disappear. We use the Distributive Property on the left side:
x*ln(5) - ln(5) = 2x*ln(3)
Gather the x terms (by subtracting x*ln(5)):
- ln(5) = 2x*ln(3) - x*ln(5)
Factor out x:
- ln(5) = x(2*ln(3) - ln(5))
Divide both sides by (2*ln(3) - ln(5)):
This is also an exact expression for the solution to your equation. This is a little more complex. But, since it uses ln, we can use our calculators to find a decimal approximation for the answer:
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