how to solve 3^(x-2) = 5^x

well, you need to get the x stuff isolated, so get rid of those exponents:

(x-2) log3 = x log5
x(log3-log5) = 2log3
x = 2log3/(log3-log5) = log9/log(3/5) = log_{_3/5}9

Thanks

To solve the equation 3^(x-2) = 5^x, we need to use logarithms. Specifically, we need to take the logarithm of both sides of the equation. Here's how you can do it step by step:

Step 1: Take the logarithm of both sides of the equation using the same base. In this case, it's common to use the natural logarithm (ln) or the logarithm base 10 (log).

Step 2: Apply the logarithm property that allows you to bring down the exponent of the variable.

Step 3: Simplify the resulting equation using the properties of logarithms.

Step 4: Solve for the variable by isolating it.

Let's solve it together:

Step 1: Taking the logarithm of both sides:
ln(3^(x-2)) = ln(5^x)

Step 2: Using the exponent property of logarithms, we bring down the exponent of the variable:
(x-2)ln(3) = xln(5)

Step 3: Simplifying the equation:
xln(3) - 2ln(3) = xln(5)

Step 4: Isolating the variable:
xln(3) - xln(5) = 2ln(3)
x(ln(3) - ln(5)) = 2ln(3)
x(ln(3/5)) = 2ln(3)

Finally, to solve for x, we divide both sides of the equation by ln(3/5):
x = 2ln(3) / ln(3/5)

Therefore, x is equal to 2 times the natural logarithm of 3 divided by the natural logarithm of 3/5.