1) Given that log2(x+1) = log4(7x-5)
a) Find possible values of X
Much appreciated
a) Find possible values of X
Much appreciated
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First lets change the base of the right hand side:
log4(7x-5) = log2(7x-5) / log2(4) = log2(7x-5) / 2log(2) = log2(7x-5) / 2
So the original ends up as 2log2(x + 1) = log2(7x - 5)
This is the same as log2(x+1)² = log2(7x - 5)
If we make each side an exponent of 2 (to undo the log2 on each side) we end up with
(x + 1)² = 7x - 5 OR x² + 2x + 1 = 7x - 5.
Collecting everything onto the left side of the equal sign we end up with
x² - 5x + 6 = 0. This is the same as (x - 2)(x - 3) = 0
Therefore x = 2 or x = 3.
log4(7x-5) = log2(7x-5) / log2(4) = log2(7x-5) / 2log(2) = log2(7x-5) / 2
So the original ends up as 2log2(x + 1) = log2(7x - 5)
This is the same as log2(x+1)² = log2(7x - 5)
If we make each side an exponent of 2 (to undo the log2 on each side) we end up with
(x + 1)² = 7x - 5 OR x² + 2x + 1 = 7x - 5.
Collecting everything onto the left side of the equal sign we end up with
x² - 5x + 6 = 0. This is the same as (x - 2)(x - 3) = 0
Therefore x = 2 or x = 3.