Properties of the Logarithm?

Question

logy = 1/4 log16 +1/2 log49

log6(base)(b^2 + 2) + log6(base)2 = 2

log3(base)(5x+5) - log3(base)(x^2 - 1) = 0

Answer 1

The first problem mainly uses the fact that a*log(b) = log(b^a) and log(a) + log(b) = log(ab).

log(y) = (1/4)log(16) + (1/2)log(49)
log(y) = log((16^(1/4))(49^(1/2))
log(y) = log(2*7)
log(y) = log(14)
y = 14

The second and third problems are more complicated, but similar to each other. I will work out the second one for you.

log_6(b^2 + 2) + log_6(2) = 2
log_6(2b^2 + 4) = log_6(36)
2b^2 + 4 = 36
2b^2 - 32 = 0
b^2 - 16 = 0
(b + 4)(b - 4) = 0
b = 4 or -4

Take note in the second line that I rewrote 2 as log_6(36) so that both sides of the equation would be in the form of log_6 functions. (For the third problem, you will probably want to rewrite 0 as log_3(1).) With log problems, it's important to always check your answers.

log_6(4^2 + 2) + log_6(2) = log_6(18) + log_6(2) = log_6(36) = 2
log_6((-4)^2 + 2) + log_6(2) = log_6(18) + log_6(2) = log_6(36) = 2

You must also work through each log operation before combining, to make sure none of the logs have invalid arguments (like a negative number) when taken individually. In this case, both answers worked out correctly.

Answer 2

1)
y>0
logy = log( 16^(1/4)*49^(1/2))=log98
y=98
2)
log6(base)2*(b^2+2)=2
2b^2+4=6^2
b^2=16
b= 4 or b=-4
3)
x>1
log3(base)(5x+5)/(x^2-1)=0
(5x+5)/(x^2-1)=1
5x+5=x^2-1
x^2-5x-6=0
x=6 or x=-1
X=6