Posted by Heather-two questions please help on Saturday, December 1, 2012 at 10:23am.
1.
let y = 6x/(3x-5)
the first step in finding the inverse is to interchange the x and y variables, so
inverse is
x = 6y/(3y-5)
3xy - 5x = 6y
3xy - 6y = 5x
y(3x - 6) = 5x
y = 5x/(3x-6) or f^-1 (x) = 5x/(3x-6)
check with a point
let x = 5 , then y = 30/(10) = 3
plug 3 in for x in the inverse
y = 15/(9-6) = 5 , looks good
2. (not even close)
I will assume you meant:
7^(2x) + 7^(x+1) - 60 = 0
7^(2x) + (7^1)(7^x) - 60 = 0
let y = 7^x
y^2 + 7y - 60 = 0
(y+12)(y-5) = 0
y = 5 or y = -12
so
7^x = 5 or 7^x = 12 , which would have no solution
if y = 5
7^x = 5
log 7^x = log5
xlog7 = log5
x = log5/log7 = appr .827
check:
LS = 7^1.654 + 7^1.827 - 60
= - .0144..
not quite zero but close enough using only 3 decimal places for x
try x = .827087475 and sub into the equation using your calculator.
can't get any closer to 0 than that.
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