A. 4 ln (2x) + ln (y) - (1/2) ln (x + 1) =
ln (2x) ^4 + ln (y) - ln (x + 1)^(1/2) =
ln 16x^4 + ln y - ln ( y) - ln √(x + 1) =
ln [ (16x^4 * y) / √(x + 1) ]
B. 2 log 5 (4x) - 1 = 11 add 1 to both sides
2 log 5 (4x) = 12 divide both sides by 2
log 5 (4x) = 6
And this says, in exponential form, that :
5^6 = 4x
15625 = 4x divide both sides by 4
3906.25 = x
C: log2 (x-6) = 5 log2 (2x)
C: log2 (x-6) = log2 (2x)^5 since the logs are same on both sides, we can forget these.....and we have......
x - 6 = (2x)^5
x - 6 = 32x^5 rewrite as
32x^5 - x + 6 = 0 using WolramAlpha.......the real solution to this equation is :
x≈-0.732151
However........this makes both logs in the original equation negative, so there is no real solution
