solve the following equation.
9^(x-1)+4*3^(x-1)=5
To solve the given equation:
1. Start by distributing the 4*3^(x-1) term. This can be done by multiplying 4 with each term inside the parentheses:
9^(x-1) + 4 * 3^(x-1) = 5
9^(x-1) + 4 * 3 * 3^(x-1) = 5
2. Simplify the equation further:
9^(x-1) + 12 * 3^(x-1) = 5
3. Next, rearrange the terms to group the similar terms together:
9^(x-1) + 12 * 3^(x-1) - 5 = 0
4. Now, we have a quadratic equation in the form of a general exponential equation:
a^2 + bx + c = 0
Let's consider 9^a as term "y". We can rewrite the equation:
y + 12y - 5 = 0
13y - 5 = 0
5. Solve the equation for y:
13y = 5
y = 5/13
6. Substitute 9^(x-1) back in for y:
9^(x-1) = 5/13
7. To get rid of the exponent, take the logarithm (base 9) of both sides:
log base 9 (9^(x-1)) = log base 9 (5/13)
8. Use the property of logarithms to simplify:
x - 1 = log base 9 (5/13)
9. Finally, isolate x by adding 1 to both sides:
x = 1 + log base 9 (5/13)
Therefore, the solution to the equation is x = 1 + log base 9 (5/13).