What is the method of solving $2^x = 2\sqrt2$?

Question

Kindly assist me in solving for $x$ in $$2^x = 2\sqrt2.$$

Answer 1

\begin{align} 2^x & = 2 \sqrt2 \\ & = 2^1 \cdot 2^{\frac{1}{2}} \\ & = 2^{1+\frac{1}{2}} \\ & = 2^{\frac{3}{2}} \\ \implies x & = \frac{3}{2} \\ \end{align}

We can do this because if the base is the same, then by index laws we can equate the index.

Hence $x=\frac32$.

Answer 2

Write: $$2^x=2^{\frac{3}{2}}$$ and you'll get it.

Answer 3

Taking log of base 2 both sides we get $$\log_{2} 2^x= \log_2 2\sqrt 2$$ $$\Rightarrow x=\log_2 2+ \log_2 \sqrt 2$$ $$\Rightarrow x=1+\frac {1}{2}=\frac {3}{2}$$

Answer 4

$$2^x = 2 \sqrt2 $$ $$2^x = 2 \times 2^ {\frac 1 2} $$ $$2^x =2^ {1+\frac 1 2} $$ $$2^x =2^ {\frac 3 2} $$ $$2^x \times 2^ {-\frac 3 2}=1 $$ $$2^ {x-\frac 3 2}=1 $$ $$2^ {x-\frac 3 2}=2^0 $$ $$ {x-\frac 3 2}=0 $$ $$ {x=\frac 3 2} $$