How to I solve 8^2x-5 = 5^x+1 using logarithm equation. Please show the steps if you can.?

8^(2x - 5) = 5^(x + 1)

log[8^(2x - 5)] = log[5^(x + 1)]

(2x - 5) log(8) = (x + 1) log(5)

2x log(8) - 5 log(8) = x log(5) + log(5)

2x log(8) - x log(5) = 5 log(8) + log(5)

x [2 log(8) - log(5)] = 5 log(8) + log(5)

x = [5 log(8) + log(5)] / [2 log(8) - log(5)]

x ≈ 4.709513175

If you mean 8^(2x - 5) = 5^(x + 1), take logs of both sides to get
(2x - 5) log 8 = (x + 1) log 5
and that is just a linear equation in x.

If you mean 8^(2x) - 5 = 5^x + 1, I don't think this can be solved analytically. You would need to use approximate numerical methods.

Maybe you mean
8^(2x - 5) = 5^(x + 1).
If you do mean that, you must NOT write it without the parentheses, as that means something completely different.

(2x - 5)*log(8) = (x+1)*log(5) =>
(2x - 5)(0.903) = (x+1)(0.699) =>
1.806 x - 4.515 = 0.699 x + 0.699 =>
1.107 x = 5.214 =>
x = about 4.7 but use a calculator.

Do you mean

8^(2x) - 5 = 5^x + 1

or maybe

(8^2) x - 5 = 5^x + 1

or maybe

8^(2x-5) = 5^x + 1

or maybe

8^(2x-5) = 5^(x+1)

?????????