How do you solve ##Log 2 + log(4x – 1) = 3##?

##x = 5/8##

You used Logarithm of a product (based on the symbol you used which is addition).

##log2 + log(4x-1) = 3## ##log 2(4x-1) = 3## ##log (8x-2) = 3## ##8x-2 = 3## ##8x = 3+2## ##8x = 5##

(then divide them with 8 so that we can remove the value of x which is not applicable), so.. ##8x-:8 = 5-:8##

##x = 5/8##