Solve for x, 9^x-1 = 3^1+x
How do I get the X out of the Exponent?
9 = 3^2 ... (3^2)^(x - 1) = 3^(2x - 2)
So.. you just mulitplied by 2?? I thought we were supposed to get x out of the exponent first before we solve? Im confused
use rules of exponents to get bases the same (3)
if bases are equal, then the exponents are equal
To solve for x in the equation 9^(x-1) = 3^(1+x), you need to get rid of the X in the exponent. One way to do this is by using the concept of logarithms. Specifically, you can take the logarithm of both sides of the equation.
By taking the logarithm, you can bring the exponent down and simplify the equation further. In this case, you can use either the logarithm base 9 or the logarithm base 3 since the equation contains both these bases.
Let's use the logarithm base 3 to solve this equation:
1. Take the logarithm base 3 of both sides of the equation:
log3 (9^(x-1)) = log3 (3^(1+x))
2. Use the logarithmic property, logb(a^c) = c * logb(a), to simplify the equation:
(x-1) * log3 (9) = (1+x) * log3 (3)
3. Evaluate the logarithms:
(x-1) * 2 = (1+x) * 1
4. Simplify the equation by distributing:
2x - 2 = x + 1
5. Rearrange the equation to isolate the x term on one side:
2x - x = 1 + 2
6. Simplify:
x = 3
Therefore, the solution to the equation 9^(x-1) = 3^(1+x) is x = 3.