solve :log (12x-10)=1+log2(4x+3)
maybe you mean to take both logs base 2? If so, then following Reiny's attempt above, that would let us pick up at
log (12x-10)=log 2 +log (8x + 6)))
log (12x-10)=log (2(8x+6))
12x-10 = 16x+12
4x = 22
x = 11/2
Possibly, but I suspect another typo.
Online, using logs to any base, say, 2, many write
log_2(x) with the underscore indicating a subscript (or log base).
Care to retype your equation?
I interpret your question as
log (12x-10)=1+log (2(4x+3))
then, replace 1 with log 10
log (12x-10)=log 10 +log (8x + 6)))
log (12x-10)=log (10(8x+6))
12x - 10 = 80x + 60
72 = -68x
x = -72/68 = -18/17
but that would make log (12x-10) undefined, so there is no solution
Let me know what you meant by log2(4x+3)
To solve the logarithmic equation log (12x - 10) = 1 + log2(4x + 3), we need to apply logarithmic rules and simplify the equation to solve for x.
Step 1: Apply the addition property of logarithms.
log (12x - 10) = log (2) + log (4x + 3)
Step 2: Apply the quotient property of logarithms.
log (12x - 10) = log ((2 * (4x + 3)) / 1)
Step 3: Set the bases equal to each other.
12x - 10 = 2 * (4x + 3)
Step 4: Distribute the 2 on the right side.
12x - 10 = 8x + 6
Step 5: Combine like terms.
12x - 8x = 6 + 10
4x = 16
Step 6: Divide both sides by 4.
x = 16 / 4
x = 4
Therefore, the solution to the equation log(12x - 10) = 1 + log2(4x + 3) is x = 4.