Question

Another log problem...

Answer 1

may i ask how you uploaded that attachment, thanks
how did you put it in an attachment

Answer 2

I got it to be (x+6)(x+4)=8
But the answer is -2

Answer 3

@perl you can attach a file.. and i had a picture of the equation

Answer 4

close you're on the right track. remember the priciples \[\log_{z} b+\log_{z} c=\log_{z} bc\] and
if \[\log_{a}b=c \] then \[a^{c}=b\]

Answer 5

pls tell me the steps that you have followed...so that i can spot the mistake?

Answer 6

log8(x + 6) = 1 – log8(x + 4)

log8(x + 6) + log8(x + 4) = 1

Using the properties of logarithms: log A + log B = log (AB)

log8[(x + 6) (x + 4)] = 1

Rewrite in exponential form:

8¹ = (x + 6) (x + 4)

x² + 10x + 24 = 8

x² + 10x + 16 = 0

(x + 2) (x + 8) = 0

x = -2 or x = -8

x = -8 is an extraneous solution, because-8 + 6 = -2, and -8 + 4 = -4, and log base 8 is undefined for negative numbers. (The domain is positive numbers only.)

So the solution is x = -2.

Verify:

log8(4) = 1 – log8(2)

log8(4) is 2/3 (Because 8^(2/3) = 2² = 4)

log8(2) is 1/3 (Because 8^(1/3) = 2)

So the above equation is equivalent to:

2/3 = 1 – 1/3

2/3=2/3