Question

Solve exponential equation: 2^x * 2^x-2= √2 (Picture included)

Answer 1

Hi there!

Th' Eqⁿ is :-

$2^x × 2^{x-2} = \sqrt{2}$

$\begin{aligned} 2^x \cdot 2^{x-2} &= \sqrt{2} \\ 2^{x + (x-2)} &= 2^{\frac{1}{2}} \\ 2^{2x - 2} &= 2^{\frac{1}{2}} \end{aligned}$

Equating the Exponent :-

Cuz' both sides of the equation have th' same base with no other terms.

$\begin{aligned} 2^{2x - 2} &= 2^{\frac{1}{2}} \\ 2x - 2 &= \tfrac{1}{2} \\ 2x &= \tfrac{1}{2} + 2 \\ 2x &= \tfrac{5}{2} \\ x &= \tfrac{5}{4} \end{aligned}$

Hence,
The required answer is x = $\frac {5}{4}$

~ Hope it helps!

Answer 2

ANSWER

x = 5/4

EXPLANATION

   $2^x \cdot 2^{x-2} = \sqrt{2}$

Note that $\sqrt{a} = a^{\frac{1}{2}}$ so $\sqrt{2} = 2^{\frac{1}{2} }$
Note that on the left-hand side, we can use exponent properties for multiplying two powers of the same base together: $a^x \cdot a^y = a^{x+y}$

   $\begin{aligned} 2^x \cdot 2^{x-2} &= \sqrt{2} \\ 2^{x + (x-2)} &= 2^{\frac{1}{2}} \\ 2^{2x - 2} &= 2^{\frac{1}{2}} \end{aligned}$

We can now equate the exponents because both sides of the equation are of the same base with no other terms.

   $\begin{aligned} 2^{2x - 2} &= 2^{\frac{1}{2}} \\ 2x - 2 &= \tfrac{1}{2} \\ 2x &= \tfrac{1}{2} + 2 \\ 2x &= \tfrac{5}{2} \\ x &= \tfrac{5}{4} \end{aligned}$

The answer is x = 5/4. We can confirm this by using this value in the original equation to get a true statement.