# Problem with Resolve

The question is: for which values of the parameter $$a$$ the equations $$4^{x+1}-3\ 2^{x+3}=2^{x+4}-64$$ and $$(a+30) \left(-9^x\right)+3^{x+4}+27^x=a \left(81-10\ 3^{x+1}\right)$$ are equivalent? Here is my unsuccessful attempt with the Resolve command.

ForAll[x, x \[Element] Reals, Equivalent[4^(x + 1) - 3*2^(x + 3) == 2^(x + 4) - 64,
27^x - (a + 30)*9^x + 3^(x + 4) == a*(81 - 10*3^(x + 1))]];Resolve[%, a, Reals]


!(* SubscriptBox["[ForAll]", RowBox[{ RowBox[{"{", "x", "}"}], ",", RowBox[{"x", "[Element]", TemplateBox[{}, "Reals"]}]}]](((-3)\ *SuperscriptBox[(2), (3 + x)] + *SuperscriptBox[(4), (1 + x)] == (-64) + *SuperscriptBox[(2), (4 + x)] [Equivalent] *SuperscriptBox[(3), (4 + x)] + *SuperscriptBox[(27), (x)] - *SuperscriptBox[(9), (x)]\ ((30 + a)) == ((81 - 10\ *SuperscriptBox[(3), (1 + x)]))\ a)))

I found the following workaround. Comparing the outputs of

Solve[4^(x + 1) - 3*2^(x + 3) == 2^(x + 4) - 64, x, Reals]


{{x->1},{x->3}}

and

Solve[27^x - (a + 30)*9^x + 3^(x + 4) == a*(81 - 10*3^(x + 1)), x, Reals]


{{x->1},{x->3},{x->ConditionalExpression[Log[a]/Log[3],a>0]}}

, one draws the conclusion that the solution is $$a\le 0$$.

Does there exist another way to this end?

• I unsuccessfully tried the SolveAlways command too. – user64494 Jan 17 at 14:35

ContourPlotof both equations
ContourPlot[{4^(x + 1) - 3*2^(x + 3) == 2^(x + 4) - 64,

shows the two equations are equivalent for a<0!
• Thank you. Unfortunately, this is rather a plausible reasoning than a proof. Also the case $a=0$ is unclear to me. – user64494 Jan 17 at 13:25
• Surely it's not a proof. But it gives an idea that the both equations might be equivalent for x==1 and x==3. – Ulrich Neumann Jan 17 at 13:56