Assistance requested solving 9^x - 6^x - 2^(2x+1) = 0 for x

[Dale10101]

Stuck

Solve for x:

\(\displaystyle \begin{array}{l}eq1:\,\,\,\,\,\,\,\,{9^x} - {6^x} - {2^{(2x + 1)}} = 0\\eq2:\,\,\,\,\,\,\,{({3^2})^x} - {[(2)(3)]^x} - {(2)^{2x}}{(2)^1} = 0\\eq3:\,\,\,\,\,\,\,{({3^x})^2} - {2^x}{3^x} - {({2^x})^2}{(2)^1} = 0\end{array}\)

\(\displaystyle \begin{array}{l}eq5:\,\,\,\,\,\,\,{\rm{Let u = }}{{\rm{3}}^x}{\rm{, v = }}{{\rm{2}}^x}\\eq6:\,\,\,\,\,\,\,{(u)^2} - uv - 2{(v)^2} = 0\end{array}\)

eq6 is a quadratic equation in two variables. I conjecture that it can be written as the product of two linear factors if the proper coefficients m and n can be found.

\(\displaystyle eq7:\,\,\,\,\,\,\,{(u)^2} - uv - 2{(v)^2}\,\,\,\mathop = \limits^? \,\,\,(u - mv)(u - nv)\)

\(\displaystyle \begin{array}{l}eq8:\,\,\,\,\,\,\,{(u)^2} - uv - 2{(v)^2}\,\,\,\mathop = \limits^? \,\,\,(u - mv)(u - nv) = {u^2} - (unv + umv) + mn{v^2}\\eq9:\,\,\,\,\,\,{(u)^2} - uv - 2{(v)^2}\,\,\,\mathop = \limits^? \,\,\,{u^2} - (n + m)uv + (mn){v^2}\end{array}\)


Equating the two middle terms and the two last terms:

\(\displaystyle eq10\,\,\,\,\,1 = (n + m),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2 = mn\)

The question mark can be removed if we can find at least one solution pair (m, n).
The text book approach for finding m and n is by trial and error and this works well if m and n are required to be integers. With practice it is easy to work out (m, n) = (-2, 1) or (m, n) = (1, -2) in both cases the result is:

\(\displaystyle eq11:\,\,\,\,\,\,\,\,{u^2} - (unv + umv) + mn{v^2} = {u^2} - ( - 2 + 1)uv + ( - 2)(1)){v^2} = {(u)^2} - uv - 2{(v)^2}\)

So eq7 becomes

\(\displaystyle eq12:\,\,\,\,\,\,\,\,\,{(u)^2} - uv - 2{(v)^2} = (u - 2v)(u + v)\)


and eq6 becomes

\(\displaystyle \begin{array}{l}eq13:\,\,\,\,\,\,\,\,\,{(u)^2} - uv - 2{(v)^2} = (u - 2v)(u + v) = 0,{\rm{ or simply,}}\\\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ }}\,\,\,\,\,\,\,(u - 2v)(u + v) = 0\end{array}\)



Which implies:

\(\displaystyle eq14:\,\,\,\,\,\,\,\,\,(u - 2v) = 0,{\rm{ and/or }}(u + v) = 0\)

Can we solve these equations for u and v and thus use


\(\displaystyle eq5:\,\,\,\,\,\,\,{\rm{Let u = }}{{\rm{3}}^x}{\rm{, v = }}{{\rm{2}}^x}\)


to yield two equations which we can solve using the inverse formula for an exponential equation?

Each equation is that of a line in 2-space, so in general 3 possibilities, they intersect at one point, they are parallel and never intersect, or they are collinear and intersect at all points on each line.

Without much mental effort it is easy to see that each line has a different slope which means they will intersect at one and only one point. It is also easy to see that they both intersect at the origin thus the solution of the two equations is (u, v) = (0, 0). This checks with eq6.

From eq5 then we can write:

\(\displaystyle eq15:\,\,\,\,\,\,\,{\rm{Let 0 = }}{{\rm{3}}^x}{\rm{, 0 = }}{{\rm{2}}^x}\)


No solution. Beep Beep, I have just been run over, Wolfram yields, without a step by step soution:

\(\displaystyle u = {3^{\frac{{LN(2)}}{{LN\left( {\frac{3}{2}} \right)}}}}\,\,\,\,\,\,\,\,\,\,\,v = {2^{\frac{{LN(2)}}{{LN\left( {\frac{3}{2}} \right)}}}}\)

How do they do that?

Thank you.

 

[Ishuda]

Stuck

Solve for x:

\(\displaystyle \begin{array}{l}eq1:\,\,\,\,\,\,\,\,{9^x} - {6^x} - {2^{(2x + 1)}} = 0\\eq2:\,\,\,\,\,\,\,{({3^2})^x} - {[(2)(3)]^x} - {(2)^{2x}}{(2)^1} = 0\\eq3:\,\,\,\,\,\,\,{({3^x})^2} - {2^x}{3^x} - {({2^x})^2}{(2)^1} = 0\end{array}\)

\(\displaystyle \begin{array}{l}eq5:\,\,\,\,\,\,\,{\rm{Let u = }}{{\rm{3}}^x}{\rm{, v = }}{{\rm{2}}^x}\\eq6:\,\,\,\,\,\,\,{(u)^2} - uv - 2{(v)^2} = 0\end{array}\)

eq6 is a quadratic equation in two variables. I conjecture that it can be written as the product of two linear factors if the proper coefficients m and n can be found.

\(\displaystyle eq7:\,\,\,\,\,\,\,{(u)^2} - uv - 2{(v)^2}\,\,\,\mathop = \limits^? \,\,\,(u - mv)(u - nv)\)

\(\displaystyle \begin{array}{l}eq8:\,\,\,\,\,\,\,{(u)^2} - uv - 2{(v)^2}\,\,\,\mathop = \limits^? \,\,\,(u - mv)(u - nv) = {u^2} - (unv + umv) + mn{v^2}\\eq9:\,\,\,\,\,\,{(u)^2} - uv - 2{(v)^2}\,\,\,\mathop = \limits^? \,\,\,{u^2} - (n + m)uv + (mn){v^2}\end{array}\)


Equating the two middle terms and the two last terms:

\(\displaystyle eq10\,\,\,\,\,1 = (n + m),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2 = mn\)

The question mark can be removed if we can find at least one solution pair (m, n).
The text book approach for finding m and n is by trial and error and this works well if m and n are required to be integers. With practice it is easy to work out (m, n) = (-2, 1) or (m, n) = (1, -2) in both cases the result is:

\(\displaystyle eq11:\,\,\,\,\,\,\,\,{u^2} - (unv + umv) + mn{v^2} = {u^2} - ( - 2 + 1)uv + ( - 2)(1)){v^2} = {(u)^2} - uv - 2{(v)^2}\)

So eq7 becomes

\(\displaystyle eq12:\,\,\,\,\,\,\,\,\,{(u)^2} - uv - 2{(v)^2} = (u - 2v)(u + v)\)


and eq6 becomes

\(\displaystyle \begin{array}{l}eq13:\,\,\,\,\,\,\,\,\,{(u)^2} - uv - 2{(v)^2} = (u - 2v)(u + v) = 0,{\rm{ or simply,}}\\\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ }}\,\,\,\,\,\,\,(u - 2v)(u + v) = 0\end{array}\)



Which implies:

\(\displaystyle eq14:\,\,\,\,\,\,\,\,\,(u - 2v) = 0,{\rm{ and/or }}(u + v) = 0\)

Can we solve these equations for u and v and thus use


\(\displaystyle eq5:\,\,\,\,\,\,\,{\rm{Let u = }}{{\rm{3}}^x}{\rm{, v = }}{{\rm{2}}^x}\)


to yield two equations which we can solve using the inverse formula for an exponential equation?

Each equation is that of a line in 2-space, so in general 3 possibilities, they intersect at one point, they are parallel and never intersect, or they are collinear and intersect at all points on each line.

Without much mental effort it is easy to see that each line has a different slope which means they will intersect at one and only one point. It is also easy to see that they both intersect at the origin thus the solution of the two equations is (u, v) = (0, 0). This checks with eq6.

From eq5 then we can write:

\(\displaystyle eq15:\,\,\,\,\,\,\,{\rm{Let 0 = }}{{\rm{3}}^x}{\rm{, 0 = }}{{\rm{2}}^x}\)


No solution. Beep Beep, I have just been run over, Wolfram yields, without a step by step soution:

\(\displaystyle u = {3^{\frac{{LN(2)}}{{LN\left( {\frac{3}{2}} \right)}}}}\,\,\,\,\,\,\,\,\,\,\,v = {2^{\frac{{LN(2)}}{{LN\left( {\frac{3}{2}} \right)}}}}\)

How do they do that?

Thank you.

I would have done basically the same as you have up to
u² - u v - 2 v² = 0.
I would then have treated that as a quadratic equation in u and got the same answers as you did
(1) u = -v or 3^x = - 2^x which doesn't admit any solutions,
or
(2) u = 2 v or 3^x = 2 2^x.
What happens if you take logs across the equation, use some of the rules for logs, rearrange the equation, solve for x, and then substitute that back in the definition for u and v?

 

[Dale10101]

Following up

I would have done basically the same as you have up to
u² - u v - 2 v² = 0.
I would then have treated that as a quadratic equation in u and got the same answers as you did
(1) u = -v or 3^x = - 2^x which doesn't admit any solutions,
or
(2) u = 2 v or 3^x = 2 2^x.
What happens if you take logs across the equation, use some of the rules for logs, rearrange the equation, solve for x, and then substitute that back in the definition for u and v?

\(\displaystyle eq14:\,\,\,\,\,\,\,\,\,(u - 2v) = 0,{\rm{\:\: and/or\:\: }}(u + v) = 0\)

\(\displaystyle eq16:\,\,\,\,\,\,\,\,\,u = 2v,{\rm{ \:\: and/or \:\: u = - v}}\)

Using substitution, \(\displaystyle eq5:\,\,\,\,\,\,\,{\rm{Let\:\: u = }}{{\rm{3}}^x}{\rm{,\:\: v = }}{{\rm{2}}^x}\)

\(\displaystyle eq:\,\,\,\,16a,\,\,\,\,\,\,{3^x} = 2({2^x}{\rm{)}}\) and/or \(\displaystyle eq:\,\,\,\,16b,\,\,\,\,\,\,{3^x} = - ({2^x}{\rm{) }}\)

By definition eq:16b has no solutions since, 3^x and 2^x are both always positive.

eq16a does relate u and v. A quick graphing shows that y₁ = 3^x and y₂ = 2^x intersectat about x = 1.7 so hopes are up for a solution.

\(\displaystyle \begin{array}{l}eq17:\,\,\,\,\,\,\,\,\,{3^x} = 2({2^x}{\rm{)}}\,\,\,\,\,\,\,{\rm{ = > }}\\{\rm{eq18:}}\,\,\,\,\,\,\,Ln({3^x}) = Ln(2({2^x}))\,\,\,\,\,\,\,\, = > \\eq19:\,\,\,\,\,\,\,\,xLn({3^{}}) = Ln(2) + xLn({2^{}}))\,\,\,\,\,\,\,\, = > \\eq20:\,\,\,\,\,\,\,\,\,\,x(Ln(3) - Ln(2)) = Ln(2)\,\,\,\,\,\,\,\,\, = > \\eq21:\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{{Ln(2)}}{{Ln\left( {\frac{3}{2}} \right)}}\end{array}\)

Therefore \(\displaystyle {9^x} - {6^x} - {2^{(2x + 1)}} = 0\) has solution when:

\(\displaystyle x = \frac{{Ln(2)}}{{Ln\left( {\frac{3}{2}} \right)}}\)

Very nice! Salamat po M. Ishuda and M. Subhotosh Khan

But now I go on to the critical step of dissecting the problem to discover why I got stuck.

After much struggle it comes down to, I believe, this: An equation defines a subset of the Cartesian product of the domains of the variables over which it is defined. Solving this equation means expanding it (a single equation) to a set of simpler equations over the same set of variables and whose union of solutions defines the same subset as the original equation.