note that you cannot simplify the right side equation log3 (4) + log3 (y+8) , by 4 + y + 8 ... they should be multiplied. recall some laws of exponents; if we are add terms of the same base, for example
log 3 + log 5
this simplifies to
log(3*5) or log 15
going back to the problem, the equation simplifies to
2log3 y = log3 4 + log3 (y+8)
log3 (y^2) = log3 (4*(y+8))
log3 (y^2) = log3 (4y + 32)
cancelling out the log3,
y^2 = 4y + 32
y^2 - 4y - 32 = 0
this is a quadratic equation. since it's factorable, let's just factor this one:
(y - 8)(y + 4) = 0
y = 8 and y = -4
note that y = -4 is extraneous, since if we substitute this back to the original, and take its log:
left side: 2log3 (-4) = undefined
thus,
y = 8
hope this helps~ :)
Solve the equation.
2log3 y = log3 4+ log3 (y+8)
Here's what ive got
take out the logs and say y^2= 4+ y +8
which simplifies to (y^2)-y=12
how do i simplfy the y?
2 answers
Jai, your a life saver! you explained everything perfectly! thank you sooo much!
that helped a ton, thanks! : )
that helped a ton, thanks! : )
1 answer