I have to solve for all real values of $x$.
$(5+2\sqrt6)^{x^2-3}+(5-2\sqrt6)^{x^2-3}=10$
I tried to take $\log_{10}$ on both sides but could not do this.
How do I do this?Thanks for any hint or answer!!
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Language nitpick: Pretty sure you don't really mean "solve for all real values of $x$" -- that would imply that for each real $x$ you have a different equation you want to solve -- but those equations wouldn't have any unknown to solve for! You probably mean that you want to solve it for $x$, which, by the way, is real. – hmakholm left over Monica May 28 '16 at 16:45
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Your comment is interesting to me. Why do we need a separate equation for each possible value? I am very interested please help me understand this on a simple level. I could open a separate question if you so direct me. Thanks much. – NoChance Aug 07 '19 at 13:21
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Hint.
note that $$(5-2\sqrt{6})=\frac{1}{5+2\sqrt{6}}$$
and use the substitution $$ \left(5+2\sqrt{6}\right)^{x^2-3x}=y $$ so the equation becomes: $$ y+\frac{1}{y}=10 $$ that becomes a second degree equation (multiply by $y \ne 0$). Solve this equation and you can find the final solution (without logarithms).
$$ y^2-10y+1=0 \quad \Rightarrow \quad y=5\pm \sqrt{25-1}=5\pm 2\sqrt{6} $$ so you have the two solutions: $$ \left(5+2\sqrt{6}\right)^{x^2-3x}=5+ 2\sqrt{6} $$ $$\left(5+2\sqrt{6}\right)^{x^2-3x}=5- 2\sqrt{6}=\left(5+ 2\sqrt{6} \right)^{-1} $$ so: $$ x^2-3x= 1 \quad \lor \quad x^2-3x= -1 $$
Now, can you find $x$?
Emilio Novati
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The last 2 equations in $x$ are contradicting and show inconsistent system. One of the answers should be $x=2$ but this can't be concluded from the above. What went wrong? – NoChance Aug 07 '19 at 13:16