SOLUTION: 3^2x+(3^x)-2=0? My professor gave us a hint and said to change out the 3x for a u..? I think that confused me. I have tried to substitute the 3x for a u. Ex: 3^2x+u-2=0 but that

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: 3^2x+(3^x)-2=0? My professor gave us a hint and said to change out the 3x for a u..? I think that confused me. I have tried to substitute the 3x for a u. Ex: 3^2x+u-2=0 but that      Log On


   

Question 212989: 3^2x+(3^x)-2=0?
My professor gave us a hint and said to change out the 3x for a u..? I think that confused me. I have tried to substitute the 3x for a u. Ex: 3^2x+u-2=0 but that does not make sense so i tried to use u(2-2)=0? I am so confused please help :)
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Here's another hint: we can rewrite 3%5E%282x%29 as 3%5E%28x%2A2%29 and %283%5Ex%29%5E2. In other words, 3%5E%282x%29=%283%5Ex%29%5E2


So 3%5E%282x%29%2B3%5Ex-2=0 transforms into %283%5Ex%29%5E2%2B3%5Ex-2=0. Now do you see how the substitution u=3%5Ex will help? If not, then just repost or ask me.


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New stuff....


3%5E%282x%29%2B3%5Ex-2=0 Start with the given equation.


%283%5Ex%29%5E2%2B3%5Ex-2=0 Rewrite 3%5E%282x%29 as %283%5Ex%29%5E2 (see above).


Now let u=3%5Ex


u%5E2%2Bu-2=0 Replace each 3%5Ex with 'u'


Take note that we now have a much simpler quadratic to solve.


Notice that the quadratic u%5E2%2Bu-2 is in the form of Au%5E2%2BBu%2BC where A=1, B=1, and C=-2


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


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


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


u+=+%28-1+%2B-+sqrt%28+1-4%281%29%28-2%29+%29%29%2F%282%281%29%29 Square 1 to get 1.


u+=+%28-1+%2B-+sqrt%28+1--8+%29%29%2F%282%281%29%29 Multiply 4%281%29%28-2%29 to get -8


u+=+%28-1+%2B-+sqrt%28+1%2B8+%29%29%2F%282%281%29%29 Rewrite sqrt%281--8%29 as sqrt%281%2B8%29


u+=+%28-1+%2B-+sqrt%28+9+%29%29%2F%282%281%29%29 Add 1 to 8 to get 9


u+=+%28-1+%2B-+sqrt%28+9+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


u+=+%28-1+%2B-+3%29%2F%282%29 Take the square root of 9 to get 3.


u+=+%28-1+%2B+3%29%2F%282%29 or u+=+%28-1+-+3%29%2F%282%29 Break up the expression.


u+=+%282%29%2F%282%29 or u+=++%28-4%29%2F%282%29 Combine like terms.


u+=+1 or u+=+-2 Simplify.


So the solutions (in terms of 'u') are u+=+1 or u+=+-2


However, we want the solutions in terms of 'x'.


Let's find the solution of 'x' that corresponds to u=1


u=3%5Ex Go back to the substitution equation.


1=3%5Ex Plug in u=1


3%5E0=3%5Ex Rewrite 1 as 3%5E0. Note: ANY number (except 0) to the 0th power is 1.


Since the bases are equal, the exponents are equal. So 0=x or x=0


So the first solution is x=0

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Let's find the solution of 'x' that corresponds to u=-2


u=3%5Ex Go back to the substitution equation.


-2=3%5Ex Plug in u=-2


log%2810%2C%28-2%29%29=log%2810%2C%283%5Ex%29%29 Take the log of both sides.


Since you CANNOT take the log of a negative number, this means that we cannot continue


So there isn't a corresponding solution of 'x' to u=-2



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Answer:


So the only solution is x=0


Check:

3%5E%282x%29%2B3%5Ex-2=0 Start with the given equation.


3%5E%282%2A0%29%2B3%5E0-2=0 Plug in x=0


3%5E0%2B3%5E0-2=0 Multiply


1%2B1-2=0 Raise 3 to the zeroth power to get 1.


2-2=0 Add


0=0 Subtract.


Since the equation is true, the solution x=0 is verified.