# How to solve this equation using logarithms?

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!!

• 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. – Henning Makholm May 28 '16 at 16:45

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 $x$?

• I have found that out too...but what to do after this? – tatan May 28 '16 at 17:05
• Added to my answer. It's all right? – Emilio Novati May 28 '16 at 17:10
• So,I get $y^2-10y+1=0$...But how do I solve $y$ from this equation? – tatan May 28 '16 at 17:25
• added another step! – Emilio Novati May 28 '16 at 19:21