Question 212989
Here's another hint: we can rewrite {{{3^(2x)}}} as {{{3^(x*2)}}} and {{{(3^x)^2}}}. In other words, {{{3^(2x)=(3^x)^2}}}



So {{{3^(2x)+3^x-2=0}}} transforms into {{{(3^x)^2+3^x-2=0}}}. Now do you see how the substitution {{{u=3^x}}} will help? <s>If not, then just repost or ask me.</s>



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New stuff....



{{{3^(2x)+3^x-2=0}}} Start with the given equation.



{{{(3^x)^2+3^x-2=0}}} Rewrite {{{3^(2x)}}} as {{{(3^x)^2}}} (see above).



Now let {{{u=3^x}}}



{{{u^2+u-2=0}}} Replace each {{{3^x}}} with 'u'



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



Notice that the quadratic {{{u^2+u-2}}} is in the form of {{{Au^2+Bu+C}}} where {{{A=1}}}, {{{B=1}}}, and {{{C=-2}}}



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



{{{u = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{u = (-(1) +- sqrt( (1)^2-4(1)(-2) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=1}}}, and {{{C=-2}}}



{{{u = (-1 +- sqrt( 1-4(1)(-2) ))/(2(1))}}} Square {{{1}}} to get {{{1}}}. 



{{{u = (-1 +- sqrt( 1--8 ))/(2(1))}}} Multiply {{{4(1)(-2)}}} to get {{{-8}}}



{{{u = (-1 +- sqrt( 1+8 ))/(2(1))}}} Rewrite {{{sqrt(1--8)}}} as {{{sqrt(1+8)}}}



{{{u = (-1 +- sqrt( 9 ))/(2(1))}}} Add {{{1}}} to {{{8}}} to get {{{9}}}



{{{u = (-1 +- sqrt( 9 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{u = (-1 +- 3)/(2)}}} Take the square root of {{{9}}} to get {{{3}}}. 



{{{u = (-1 + 3)/(2)}}} or {{{u = (-1 - 3)/(2)}}} Break up the expression. 



{{{u = (2)/(2)}}} or {{{u =  (-4)/(2)}}} 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^x}}} Go back to the substitution equation.



{{{1=3^x}}} Plug in {{{u=1}}}



{{{3^0=3^x}}} Rewrite {{{1}}} as {{{3^0}}}. 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^x}}} Go back to the substitution equation.



{{{-2=3^x}}} Plug in {{{u=-2}}}



{{{log(10,(-2))=log(10,(3^x))}}} 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^(2x)+3^x-2=0}}} Start with the given equation.



{{{3^(2*0)+3^0-2=0}}} Plug in {{{x=0}}}



{{{3^0+3^0-2=0}}} Multiply



{{{1+1-2=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.