Posted by Heathertwo questions please help on Saturday, December 1, 2012 at 10:23am.
1. the function f(x)=6x/3x5 is one to one. find its inverse and check your answer.
And on the second one can you tell me if I am correct.
2. solve the following exponential solution 7^2x+7^x+160=0
7^2 x+7^x59=0
(49x)+7^x59=0
49x/49=0/49
x=0/49
x=0

algebra  Reiny, Saturday, December 1, 2012 at 11:35am
1.
let y = 6x/(3x5)
the first step in finding the inverse is to interchange the x and y variables, so
inverse is
x = 6y/(3y5)
3xy  5x = 6y
3xy  6y = 5x
y(3x  6) = 5x
y = 5x/(3x6) or f^1 (x) = 5x/(3x6)
check with a point
let x = 5 , then y = 30/(10) = 3
plug 3 in for x in the inverse
y = 15/(96) = 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)(y5) = 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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