SOLUTION: 1.) Solve each equation. a.) log7(2x)^2+log7 (3)=2 b.) log4 (8)+log4 (x^2-8)=8 c.) log12 (-3a-8)+log12 (a+4) d.) ln-3k=ln(-2k-4)

Algebra ->  Trigonometry-basics -> SOLUTION: 1.) Solve each equation. a.) log7(2x)^2+log7 (3)=2 b.) log4 (8)+log4 (x^2-8)=8 c.) log12 (-3a-8)+log12 (a+4) d.) ln-3k=ln(-2k-4)       Log On


   

Question 690972: 1.) Solve each equation.
a.) log7(2x)^2+log7 (3)=2
b.) log4 (8)+log4 (x^2-8)=8
c.) log12 (-3a-8)+log12 (a+4)
d.) ln-3k=ln(-2k-4)
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
1.) Solve each equation.
a.) log7(2x)^2+log7 (3)=2
log7[(2x)^2*3]=2
convert to exponential form
7^2=4x^2*3=12x^2
12x^2=49
x^2=49/12
x=±√(49/12)≈±2.02
..
b.) log4 (8)+log4 (x^2-8)=8
log4[8*(x^2-8)]=8
4^8=8*(x^2-8)=8x^2-64
8x^2-64=65536
8x^2=65536+64=65600
x^2=8200
x≈±90.55
..
c.) log12 (-3a-8)+log12 (a+4)
log12[(-3a-8)(a+4)]=log12[(-3a^2-20a-32)]=log12[-(3x+8)(x+4)]
..
d.) ln-3k)=ln(-2k-4)=?
ln argument mssing