Solve the equation : 8^2x-1 = 16^x+1

i know you have to log both sides

log8^2x-1 = log16^x+1

and then you use the exponential rule

2x-1 log8 = x+1 log16

but then i don't know what to do...

Your steps are correct if the equation was:

8^(2x-1) = 16^(x+1)
and assuming this is the case.

Since 8=2^3, and 16=2^4,
you can simplify further using the same rule of logarithms:
log(a^b)=b×log(a)

To solve the equation 8^(2x-1) = 16^(x+1), you are correct that taking the logarithm of both sides is a good first step. Here's what you need to do next:

1. Apply the logarithmic property log(a^b) = b * log(a) to both sides of the equation. This yields:
(2x - 1) * log(8) = (x + 1) * log(16)

2. Recall that log(8) and log(16) can be simplified. Use the change of base formula to change the base of the logarithm, for example, to natural logarithm (log base e), or to logarithm base 10. Either choice will work, but let's use base 10 logarithm for this explanation.

3. Rewrite log(8) and log(16) in terms of log base 10:
log(8) = log(2^3) = 3 * log(2), and
log(16) = log(2^4) = 4 * log(2).

4. Substitute these values back into the equation, giving:
(2x - 1) * (3 * log(2)) = (x + 1) * (4 * log(2))

5. Simplify both sides:
6x - 3log(2) = 4x + 4log(2)

6. Move all terms with x to one side and all terms with log(2) to the other side:
6x - 4x = 4log(2) + 3log(2), which becomes:
2x = 7log(2)

7. Divide both sides by 2 to solve for x:
x = (7log(2)) / 2

8. Finally, evaluate the right side of the equation using a calculator to find the numerical value for x.