SOLUTION: find the exact value(s) of x such that 9^x-1-3^x-1-2=0

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Question 214854: find the exact value(s) of x such that 9^x-1-3^x-1-2=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
9%5E%28x-1%29-3%5E%28x-1%29-2=0 Start with the given equation.


%289%5Ex%29%2F9%5E1-%283%5Ex%29%2F3%5E1-2=0 Rewrite the left side using the identity x%5E%28y-z%29=%28x%5E%28y%29%29%2F%28x%5E%28z%29%29


%289%5Ex%29%2F9-%283%5Ex%29%2F3-2=0 Simplify


Multiply EVERY term by the LCD 9 to clear out the fractions.


9%5Ex-3%2A3%5Ex-18=0 Simplify


%283%5E2%29%5Ex-3%2A3%5Ex-18=0 Rewrite 9 as 3%5E2


3%5E%282x%29-3%2A3%5Ex-18=0 Multiply the exponents.


%283%5Ex%29%5E2-3%2A3%5Ex-18=0 Factor out the exponent 2 using the identity x%5E%28y%2Az%29=%28x%5E%28z%29%29%5Ey


Now let's make a substitution to make things easier on us. Let z=3%5Ex (since there are 2 copies of 3%5Ex)


z%5E2-3z-18=0 Replace each 3%5Ex with 'z'


Notice that the quadratic z%5E2-3z-18 is in the form of Az%5E2%2BBz%2BC where A=1, B=-3, and C=-18


Let's use the quadratic formula to solve for "z":


z+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


z+=+%28-%28-3%29+%2B-+sqrt%28+%28-3%29%5E2-4%281%29%28-18%29+%29%29%2F%282%281%29%29 Plug in A=1, B=-3, and C=-18


z+=+%283+%2B-+sqrt%28+%28-3%29%5E2-4%281%29%28-18%29+%29%29%2F%282%281%29%29 Negate -3 to get 3.


z+=+%283+%2B-+sqrt%28+9-4%281%29%28-18%29+%29%29%2F%282%281%29%29 Square -3 to get 9.


z+=+%283+%2B-+sqrt%28+9--72+%29%29%2F%282%281%29%29 Multiply 4%281%29%28-18%29 to get -72


z+=+%283+%2B-+sqrt%28+9%2B72+%29%29%2F%282%281%29%29 Rewrite sqrt%289--72%29 as sqrt%289%2B72%29


z+=+%283+%2B-+sqrt%28+81+%29%29%2F%282%281%29%29 Add 9 to 72 to get 81


z+=+%283+%2B-+sqrt%28+81+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


z+=+%283+%2B-+9%29%2F%282%29 Take the square root of 81 to get 9.


z+=+%283+%2B+9%29%2F%282%29 or z+=+%283+-+9%29%2F%282%29 Break up the expression.


z+=+%2812%29%2F%282%29 or z+=++%28-6%29%2F%282%29 Combine like terms.


z+=+6 or z+=+-3 Simplify.


So the solutions are z+=+6 or z+=+-3


Recall that we let z=3%5Ex. Since 3%5Ex%3E0 for all 'x', the solution z=-3 isn't possible (since 3%5Ex%3C%3E-3)

So we're only going to focus on z+=+6


z=3%5Ex Start with the given equation.


6=3%5Ex Plug in z+=+6


log%283%2C%286%29%29=x Convert to logarithmic form.


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Answer:


So the only solution is x=log%283%2C%286%29%29