Question
Question: How do you solve \[{4^x} - {2^x} = 0\] ?...
How do you solve 4x−2x=0 ?
Solution
Hint : Here we have to solve the above equation. The above equation is in the form of exponential form. The exponential number is defined as the number of times we multiply the number by itself. So, we can’t solve this directly so we apply log to it and solve further.
Complete step-by-step answer :
The logarithmic function and the exponential function are both inverse of each other. The exponential number can be written in the form of a logarithmic number and likewise we can write the logarithmic number in the form of an exponential number.
Now we have equation 4x−2x=0
This can be written as 4x=2x
Apply log on the both sides we have
By the logarithmic property we have logam=mloga , using this property we have
⇒xlog4=xlog2
For any values of x other than zero the log4 is not equal to log2. Therefore, we have
x=0 .
Hence, we have solved the equation and determined the value of x.
We can also solve by another method.
So now consider
4x−2x=0
Take 2x to the RHS so we have
⇒4x=2x
The number 4 can be written in the form of exponent. As we know that square of 2 is 4. Soe we have
⇒(22)x=2x
By the property of the logarithmic function the above inequality is written as
⇒22x=2x
In the above equation the base are same, so we can equate the exponent
So we have
⇒2x=x
Take x to LHS we have
⇒2x−x=0
On simplifying we get
⇒x=0
Hence we have solved the given equation.
So, the correct answer is “x = 0”.
Note : The exponential number is inverse of logarithmic. But here we have not used this. We have applied the log on both terms. The logarithmic functions have several properties on addition, subtraction, multiplication, division and exponent. So we have to use logarithmic properties