Exercise
Solve the equation:
4^{x-1}+2^{x-2}=68
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
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