1) log(sqrt)x^3-17=1/2 2)ln(1-x)+ln(3-x)=ln8?

Question

(Exponential and logarithmic equations)Solve for x Algebraically.

Answer 1

#1: I'm going to assume you mean log (√x^3) - 17 = 1/2, so if that's not what you mean, this answer is wrong.

log (√x^3) - 17 = 1/2
log (x^(3/2)) = 35/2
3/2 log x = 35/2
log x = 35/3
x=10^(35/3)

#2: ln (1-x)+ln(3-x)=ln 8
ln ((1-x)(3-x))=ln 8
(1-x)(3-x)=8
x²-4x+3=8
x²-4x-5=0
(x-5)(x+1)=0

x=5 or x=-1, however x≠5 since that would mean 1-x<0, so ln (1-x) would not exist. Therefore, x=-1.

Answer 2

1. You can clean this up a bit to get
3/2 log x = 17.5 and solve that, but I wonder if the 17 is also in the log term.

2. You can raise each side to the "eth" power to get (1-x)(3-x)=8. This gives you a quadradic which you can solve "the usual way".

Answer 3

1) Use log rules:
log(sqrt(x^3))-17=1/2
log(x^(3/2))-17=1/2
(3/2)log(x)-17=1/2
(3/2)log(x)=17.5
log(x)=11.66
x=4.64158 times 10 to the 11th

2)
ln(1-x)+ln(3-x)=ln(8)
ln((1-x)/(3-x))=ln(8)
(1-x)/(3-x)=8
1-x=8*(3-x)
1-x=24-8x
-23=-7x
x=3.2857