Question

If $(2^x–4{)}^3+(4^x–2{)}^3=(4^x+2^x–6{)}^3$, then how many value(s) of x are possible?

• A. 3
• B. 0
• C. 2
• D. 1

Answer 1

The correct option is A 3

Given: $(2^x–4{)}^3+(4^x–2{)}^3=(4^x+2^x–6{)}^3$
Using the identity $(x+y{)}^3=x^3+y^3+3xy(x+y)$:
$(4^x+2^x–6{)}^3=[(2^x–4)+(4^x–2){]}^3=(2^x–4{)}^3+(4^x–2{)}^3+3(2^x–4)(4^x–2)(2^x–4+4^x–2)$
$\Rightarrow (2^x+4^x–6{)}^3=(4^x+2^x–6{)}^3+3(2^x–4)(4^x–2)(2^x+4^x–6)$ (Given)
$\Rightarrow 3(2^x–4)(4^x–2)(2^x+4^x–6)=0$
$\Rightarrow 2^x-4=0,4^x-2=0$ or $2^x+4^x-6=0$
Consider $2^x–4=0.$
$\therefore 2^x=4$
$\Rightarrow 2^x=2^2$
$\Rightarrow x=2$
Now, consider $4^x–2=0$.
$\therefore 4^x=2$
$\Rightarrow 2^{2x}=2^2$
$\Rightarrow 2x=1$
$\Rightarrow x=\frac{1}{2}$
Now, consider $2^x+4^x–6=0.$
$\Rightarrow 2^x+2^{2x}-6=0$
$\Rightarrow 2^{2x}+(3-2)2^x-6=0$
$\Rightarrow (2^x{)}^2+3\times 2^x-2\times 2^x-6=0$
$\Rightarrow 2^x(2^x+3)-2(2^x+3)=0$
$\Rightarrow (2^x-2)(2^x+3)=0$
$\Rightarrow 2^x–2=0$ or $2^x+3=0$
$\Rightarrow 2^x=2$ or $2^x=–3$
Since the value of $2^x$ cannot be negative.
$\Rightarrow 2^x=2$
$\Rightarrow 2^x=2^1$
$\Rightarrow x=1$
Thus, there are three possible values of x for the given expression.
Hence, the correct answer is option (1).