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?
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~ :)
Jai, your a life saver! you explained everything perfectly! thank you sooo much!
that helped a ton, thanks! : )
To simplify the equation further, you can rearrange it and then solve for y using the quadratic formula or factoring. Let's solve it using factoring:
Start with the equation:
y^2 - y = 12
Rearrange it to set it equal to zero:
y^2 - y - 12 = 0
Now, we need to factor the equation:
(y - 4)(y + 3) = 0
Set each factor equal to zero and solve for y:
y - 4 = 0 or y + 3 = 0
For y - 4 = 0, you add 4 to both sides:
y = 4
For y + 3 = 0, you subtract 3 from both sides:
y = -3
So the possible solutions to the equation are y = 4 or y = -3.
To simplify the equation (y^2) - y = 12, you need to rearrange the equation and bring all the terms to one side to have it in the form of a quadratic equation.
Start by subtracting 12 from both sides of the equation:
(y^2) - y - 12 = 0
Now, you have a quadratic equation of the form ax^2 + bx + c = 0, where a = 1, b = -1, and c = -12.
To solve this quadratic equation, you can use factoring, completing the square, or the quadratic formula. Since it may not be immediately apparent how to factor this equation, let's use the quadratic formula:
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, the equation is (y^2) - y - 12 = 0, so applying the quadratic formula, we get:
y = (-(-1) ± √((-1)^2 - 4(1)(-12))) / (2(1))
Simplifying further:
y = (1 ± √(1 + 48)) / 2
y = (1 ± √49) / 2
y = (1 ± 7) / 2
Hence, the solutions to the equation are:
y = (1 + 7) / 2 = 8 / 2 = 4
y = (1 - 7) / 2 = -6 / 2 = -3
Therefore, the solutions to the given equation are y = 4 and y = -3.