fbpx
calculus online - Free exercises and solutions to help you succeed!

Equations – Solving an exponential equation – Exercise 1687

Exercise

Solve the equation:

4^{x-1}+2^{x-2}=68

Final Answer


x=4

Solution

4^{x-1}+2^{x-2}=68

In order to reach the same base in both expressions, we define a new variable. To do this, we will use the Power Rules:

2^{2(x-1)}+2^{x-1-1}=68

{(2^{x-1})}^2+2^{x-1}\cdot 2^{-1}=68

Now, we define a new variable:

y=2^{x-1}

We set the new variable:

y^2+y\cdot\frac{1}{2}=68

It is a quadratic equation. We rearrange its expressions:

y^2+\frac{1}{2} y-68=0

We multiply the equation by 2 to get rid of fractions:

2y^2+y-136=0

The coefficients of the equation are

a=2, b=1, c=-136

We solve it with the quadratic formula. Putting the coefficients in the formula gives us

y_{1,2}=\frac{-1\pm \sqrt{1^2-4\cdot 2\cdot (-136)}}{2\cdot 2}=

=\frac{-1\pm \sqrt{1089}}{4}=

=\frac{-1\pm 33}{4}

Hence, we get the solutions:

y_1=\frac{-1+ 33}{4}=\frac{32}{4}=8

y_2=\frac{-1- 33}{4}=\frac{-34}{4}=-8.5

We go back to the original variable. From the first solution we get

y=8, y=2^{x-1}

8=2^{x-1}

2^3=2^{x-1}

Since the bases are equal, the powers are equal as well:

3=x-1

4=x

From the second solution we get

y=-8.5, y=2^{x-1}

-8.5=2^{x-1}

This equation has no real solution, because the left side is negative, while the right side is positive for every x.

Hence, the only solution of the equation is

x=4

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! 

Share with Friends

Leave a Reply