Equations – Solving an exponential equation – Exercise 1687

Exercise

Solve the equation:

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

Final Answer

Show final answer

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!