How Do You Solve Log 2 Log 4x 1 3

/in /by

##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##

 
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