Ab asked in Science & MathematicsMathematics · 1 month ago

# URGENT help needed with maths question ?

Find all values of y such that:

log2(3y + 2)  +  log2(4y - 4)  = 3

The 2's next to the logs are sub scripts.

Can anyone solve this with good working?

Thanks

Relevance
• 1 month ago

If your logarithms have the same base, then:

log(a) + log(b) = log(a * b)

log(a) - log(b) = log(a/b)

log[2](3y + 2) + log[2](4y - 4) = 3

log[2]((3y + 2) * (4y - 4)) = 3

(3y + 2) * (4y - 4) = 2^3

4 * (3y + 2) * (y - 1) = 8

(3y + 2) * (y - 1) = 2

3y^2 - 3y + 2y - 2 = 2

3y^2 - y - 4 = 0

y = (1 +/- sqrt(1 + 48)) / 6

y = (1 +/- 7) / 6

y = 8/6 , -6/6

y = 4/3 , -1

Test each answer in the original problem.  One or both might be extraneous.

log[2](3 * (4/3) + 2) + log[2](4 * (4/3) - 4) =>

log[2](4 + 2) + log[2](16/3 - 12/3) =>

log[2](6) + log[2](4/3) =>

log[2](6 * (4/3)) =>

log[2](8) =>

log[2](2^3) =>

3 * log[2](2) =>

3 * 1 =>

3

log[2](3 * (-1) + 2) + log[2](4 * (-1) - 4) =>

log[2](-3 + 2) + log[2](-4 - 4) =>

log[2](-1) + log[2](-8)

And here's where we have a problem.  log[2](-1) isn't defined with real numbers.  It's complex.  If we're permitted to use complex values, then y = -1 is a valid solution.  Otherwise, it's extraneous.

y = 4/3

• Ab1 month agoReport

Thank you lots!!

• 1 month ago

raise 2 by the equation so that 2^(log2(x)) and not that log2(x) + log2(y) = log2(x*y)

(3y +2)(4y - 4) = 2^3 = 8

12y^2 +8y - 12y -8 = 8 --> 12y^2 - 4y = 0 ---> 3y - 4 = 0 --> y = 4/3

• Ab1 month agoReport

Thank you :)