Hey..can u guys help me out with these numbers??

Question

1. (3^x) / (3^x + 3) = 1/3

2. log x+2 (3x^2 + 4x - 14) = 2

p.s: x + 2 is the base to which u r finding the logarithm.

hey..the answer to the question is 0.37..but i dont have any idea as to have it is obtained..anybody else out there??

Answer 1

2.) log (x + 2) (3x^2 + 4x - 14) = 2

Step 1: Use the change of base identity and simplify.
log (a) b = log b / log a
log (x + 2) (3x^2 + 4x - 14) = log (3x^2 + 4x - 14) / log (x + 2) = 2
log (3x^2 + 4x - 14) / log (x + 2) = 2

Step 2: Move the denominator to the other side and distribute into the logarithm.
log (3x^2 + 4x - 14) = 2 log (x + 2)
log (3x^2 + 4x - 14) = log (x + 2)^2
log (3x^2 + 4x - 14) = log (x^2 + 4x + 4)

Step 2: Remove log from both sides and solve.
log (3x^2 + 4x - 14) = log (x^2 + 4x + 4)
3x^2 + 4x - 14 = x^2 + 4x + 4
3x^2 + 4x - 14 - (x^2 + 4x + 4) = 0
2x^2 - 18 = 0
2x^2 = 18
x^2 = 9
x = 3 (solution!)

Answer 2

1.
=> 3^(x+1) - 3^x = 3
=> (3^x)(3-1) = 3
=> (3^x) = 3/2
apply logarithm based 3:
x = log_3_(3/2) which is somewhere in (0 ; 1)

edit: here`s how: x = log_3_(3/2) = 0.37