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October 1, 2014

October 1, 2014

Posted by **Heather-two questions please help** on Saturday, December 1, 2012 at 10:23am.

And on the second one can you tell me if I am correct.

2. solve the following exponential solution 7^2x+7^x+1-60=0

7^2 x+7^x-59=0

(49x)+7^x-59=0

49x/49=0/49

x=0/49

x=0

- algebra -
**Reiny**, Saturday, December 1, 2012 at 11:35am1.

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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