Question

How do you solve ${{2}^{x}}=0.5$ ?

Answer 1

Hint: Firstly, write 0.5 in fraction form and then simplify in such a way that it remains in the fraction form itself. Now write in terms of powers of 2 because it will be easier to cancel the base which is 2 and then equate the powers to find the value of x.

Complete step-by-step solution:
The given expression is, ${{2}^{x}}=0.5$
Now write 0.5 in fraction form.
Always check the number of decimal places (number of digits) after the decimal place to convert the decimal into a fraction.
The number of digits is then written to the power of $10\;$ and then placed in the denominator and in the numerator, the number will be written the same as it is, but without the decimal point.
We can write it as, $\dfrac{5}{10}$
Now simplify the fraction.
$\Rightarrow {{2}^{x}}=\dfrac{5}{2\times 5}$
$\Rightarrow {{2}^{x}}=\dfrac{1}{2}$
Now,
We can write $\dfrac{1}{2}={{2}^{-1}}$
Upon substituting this we get,
$\Rightarrow {{2}^{x}}={{2}^{-1}}$
Now since the bases are the same, equate the powers.
Upon equating the powers, we get,
$\Rightarrow x=-1$
Hence the value of x in ${{2}^{x}}=0.5$ is -1.

Note: There is another method to solve this question.
The logarithm of a given constant $y$ is the exponent to which another fixed constant, the base $b$ , must be raised, to produce that constant $y$.
$\Rightarrow {{\log }_{a}}(b)=y$
Can be written as, $b={{a}^{y}}$
According to our question, the reverse is given.
${{2}^{x}}=0.5$
Here $a=2,y=x,b=0.5$
Now write this in logarithmic form.
We get,
$\Rightarrow {{\log }_{2}}0.5=x$
Now we can write $\dfrac{1}{2}={{2}^{-1}}$
Upon substituting that we get,
$\Rightarrow {{\log }_{2}}{{2}^{-1}}=x$
We know that according to the Law of powers of logarithms if we have the function, $f(x)={{\log }_{a}}({{b}^{c}})$ .Then we can convert into power form as, $f(x)=c{{\log }_{a}}(b)$ .
Hence, we can evaluate as,
$\Rightarrow -1{{\log }_{2}}2=x$
If the base and the logarithm value is the same, they cancel out to get $1$
$\Rightarrow {{\log }_{2}}2=1$
Hence, x=-1.