Solve for x

This looks difficult at first, but notice that the numbers look like part of a trigonometric equation. Rewrite the equation, trying to achieve some recognizable trigonometric value. $\begin{aligned} \left(\sqrt{3}+1\right)^{2x}+\left(\sqrt{3}-1\right)^{2x}&=2^{3x}\\ \left(\sqrt{3}+1\right)^{2x}+\left(\sqrt{3}-1\right)^{2x}&=\left(2\sqrt{2}\right)^{2x}\\ \left(\dfrac{\sqrt{3}+1}{2\sqrt{2}}\right)^{2x}+\left(\dfrac{\sqrt{3}-1}{2\sqrt{2}}\right)^{2x}&=1\\ \left(\dfrac{\sqrt{6}+\sqrt{2}}{4}\right)^{2x}+\left(\dfrac{\sqrt{6}-\sqrt{2}}{4}\right)^{2x}&=1\\ \left(\sin\dfrac{5\pi}{12}\right)^{2x}+\left(\cos\dfrac{5\pi}{12}\right)^{2x}&=1 \end{aligned}$ You can solve that by invoking the identity $\sin^2\theta+\cos^2\theta=1,$ seeing that $x=1$ is the only solution. Thus, there is $\boxed{1}$ solution.

$\frac{(\sqrt{3}+1)^{2x}}{2^{3x}}$ and $\frac{(\sqrt{3}-1)^{2x}}{2^{3x}}$ as functions of x are 2 strictly monotonically decreasing functions whose sum is again a striclty monotonically decreasing function which evaluates to $2$ at $x = 0$ and $0$ as x tends to infinity. Thus it must pass through 1 exactly 1 times. So the number of solutions is also equal to 1

taking log both sides, it will be very easy to see that only x=0 satisfies that log eq. Hence no. of solution is 1.

Its js d guessing thing,see 2x is the index,fr instance remove 'x' & solve.so The final ans comes 8 & 2^3=8,remove x! So both sides come same! Means x=1!

by Fermat's theorem,value of x ranges only from 0 to 1 it cannot be greater than 2, it follows x cannot also be zero, only 1 could be the value of x that only statisfies the given equation.

${ \left( \sqrt { 3 } +1 \right) }^{ 2x }+{ \left( \sqrt { 3 } -1 \right) }^{ 2x }={ 2 }^{ 3x }$

$\Rightarrow \quad { \left( \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \right) }^{ 2x }+\left( \frac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } } \right) ^{ 2x }=1$

$\Rightarrow \quad \left( \sin { 75 } \right) ^{ 2x }+\left( \cos { 75 } \right) ^{ 2x }=1\\ \Rightarrow \quad x\quad must\quad be\quad equal\quad to\quad 1$

So, the total number of solutions =1

simply u can check by putting, x=0,1,2......only x=1 satisfies th equ., so no. of values is 1.

x cannot be greater than 2 to contradict fermat's last theorem .....and x=2 do not satisfy the equation

Oh god! I guess there's an easier solution. Take the terms to be like a2+b2,

add and subtract 2ab, to get (a+b)2-2ab=2^3x

simplification leads to 1

2^x-2^(x+1)=2^3x

or 6^x=2^2x+2

Only soln is 1!

(sqtr(3) + 1)^2 = 4 + 2sqtr(3).........(sqtr(3) - 1)^2 = 4 - 2sqtr()........2^3 = 8...
Therefore eqution is....> [4 + 2sqtr(3)]^x + [ 4 - 2sqtr(3)]^x = 8^x...........

Since[ 4 + 2sqtr(3)] + [ 4 - 2sqtr(3)] = 8, ....x = 1.

The most simple way:. { 2 }^{ 3x } = { ( }{ 2 }^{ 3 })^{ x } or { (2 }^{ x })^{ 3 }

So: (\sqrt { { 3 }^{ x } } +\quad 1)^{ 2 }\quad +\quad (\sqrt { { 3 }^{ x } } -\quad 1)^{ 2 }\quad =\quad { (2 }^{ 3 })^{ x }

Then: { 3 }^{ x }\quad +\quad 2\sqrt { { 3 }^{ x } } +\quad 1\quad +\quad { 3 }^{ x }\quad -\quad 2\sqrt { { 3 }^{ x } } +\quad 1\quad =\quad { 8 }^{ x }

Now: { 3 }^{ x }\quad +\quad { 3 }^{ x }\quad +\quad 2\quad =\quad { 8 }^{ x }

Just looking at this one (Now), we can afirm that x = 1. If you try x = 2 or bigger, it'll always be wrong. I know that isn't the right way to do this, but, if you don't have the necessary skills, try this way! That's the brazillian way ;)

if x=1,LHS=4+4=8