Pre-Calc: Logarithms?

Question

Solve analytically.

1) log(2x+3) = -log(3-x)

2) 0.16^(4+3x) = 1.3^(8-x)

3) 3log(x^3) = log(x-1)

4) 12^(0.5x) = 8^(0.1x-4)

*All logs are base of 10.

Answer 1

All logs are base 10.

1) log(2x + 3) = -log(3 - x)
log(2x + 3) + log(3 - x) = 0
log[(2x + 3)(3 - x)] = 0
log(9 + 3x - 2x²) = 0
9 + 3x - 2x² = 10^0 = 1
8 + 3x - 2x² = 0
2x² - 3x - 8 = 0

x = {3 ± √[3² - 4*2*(-8)]} / (2*2)
x = {3 ± √[9 + 64]} / 4
x = (3 ± √73) / 4
____________

2) 0.16^(4 + 3x) = 1.3^(8 - x)
log[0.16^(4 + 3x)] = log[1.3^(8 - x)]
(4 + 3x)log(0.16) = (8 - x)log(1.3)
(3x)log(0.16) + xlog(1.3) = 8log(1.3) - 4log(0.16)
[3log(0.16) + log(1.3)]x = 8log(1.3) - 4log(0.16)

x = [8log(1.3) - 4log(0.16)] / [3log(0.16) + log(1.3)]
__________

3) 3log(x^3) = log(x - 1)
log(x^(3*3)) = log(x - 1)
log(x^9) = log(x - 1)
x^9 = x - 1
x^9 - x + 1 = 0

By Descartes' Rule of Signs

f(x) = +x^9 - x + 1
has 2 sign changes so there are 2 or 0 real positive roots

f(-x) = (-x)^9 - (-x) + 1 = -x^9 + x + 1
has 1 sign change so there is exactly 1 real negative root

Obviously x = 0 is not a root. So there are either 3 or 1 real roots. As it happens there is exactly one real root and therefore it must be negative. With software I find that:

x ≈ -1.08507

This is the only real solution.
__________

4) 12^(0.5x) = 8^(0.1x - 4)
log[12^(0.5x)] = log[8^(0.1x - 4)]
(0.5x)log(12) = (0.1x - 4)log(8)
(x/2)log(12) - (x/10)log(8) = -4log(8)
(x/10)[5log(12) - log(8)] = -4log(8)
x[5log(12) - log(8)] = -40log(8)

x = -40log(8) / [5log(12) - log(8)]
___________

Answer 2

let's try some ...

1) log(2x+3) = -log(3-x), gather
log(2x+3) + log(3-x) = 0, prod rule
log[(2x+3)(3-x)] = 0, so
10^0 = (2x+3)(3-x), def of log, or
1 = (2x+3)(3-x), expand/ distribute
1 = 3(2x+3)- x (2x+3) or
1 = 6x + 9 - 2x^2 - 3x, or
1 = 3x + 9 - 2x^2, gather to one side
1 - 3x - 9 + 2x^2 = 0, rearrange
2x^2 - 3x - 8 = 0, solve by quadratic formula

2) 0.16^(4+3x) = 1.3^(8-x), take logs of both sides
log[ 0.16^(4+3x) ]= log [1.3^(8-x)]
(4+3x) log(.16) = (8-x) log(1.3), cross multiply
(4+3x)/ (8-x) = log(1.3)/ log(.16) , or
(4+3x)/ (8-x) = - .14316, cross mult
(4+3x) = - .14316(8-x), solve for x.

3) 3log(x^3) = log(x-1), power rule
log(x^3)^3 = log(x-1), or
log(x^9) = log(x-1), implying
x^9 = x - 1 or
x^9 -x + 1 = 0, Solve this graphically, I suggest.

4) take logs of both sides