# Equations – Solving an exponential equation – Exercise 1687

Exercise

Solve the equation:

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

$$x=4$$

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

$$x=4$$

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