Exponential Equations

🏆Practice exponential equations

Exponential Equations

Exponential equations are equations in which the unknown appears with power, such as ax=bxa^x=b^x.
When we encounter an exponential equation, we will try to solve it by the most appropriate method.

Solving Exponential Equations:

There are two main ways to solve an exponential equations:

Identical Bases Method:

The method is based on the principle that says when 
ax=aya^x=a^y
Then 
x=yx=y

That is, if we reach a situation where the bases are identical, we can easily compare the powers and find the unknown we are looking for.
As such, to use this method we need to get both sides of the equation to have the same base.
For example, in 2x=82x=8, since 88 can be written as 232^3, we set the exponents equal and solve: x=3x = 3.

Demonstration of solving exponential equations using the method of identical bases.

Quadratic Equation Method (Substitution of t):

For more complex equations, like a2x+ax=ba^{2x} + a^x = b, use substitution. Let t=axt = a^x, turning the equation into a quadratic form. Solve for tt, then back-substitute to find the value of xx.

Solving exponential equations using the quadratic equation method and substituting

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Test yourself on exponential equations!

einstein

Solve for x:

\( 3^29^2=3^x \)

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Identical Bases Method

The method is based on the principle that says when 
ax=aya^x=a^y
Then 
x=yx=y

That is, if we reach a situation where the bases are identical, we can easily compare the powers and find the unknown we are looking for.
If we could not get to the same bases, we will move on to the second method:


Quadratic Equation Method and Substitution of t.

In this method, we use substitution to replace elements that are cumbersome to work with, with elements we know how to handle.
For this method to work, we must bring the function to a specific form.
We will have to bring our exercise to a state where there is an element, the same quadratic element, and a free number (without variables).
Once we reach this state, we will use a power player tt and place it in place of the element we have with a variable power.
Thus, we will arrive at a common quadratic equation that we can easily solve.

If we could not obtain the same bases, we will resort to the second method:
Solving exponential equations using a quadratic equation and placing tt

Thus, we will have to bring our exercise to a state in which there is an element, the same squared, and a free number (without variable).
(Ax)2+B×Ax+C=0(A^x )^2+B\times A^x+C=0
Once we reach this state, we will use a booster player tt and place it in place of the element we have with a variable power.
t2+Bt+C=0t^2+Bt+C=0
In this way, we will obtain a standard quadratic equation that we can easily solve.
Important! The solution of the quadratic equation is not the final result of the exercise.
We will not forget to place the values of the tt  we found to find the variable values we are looking for xx.


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Identical Bases Method - Example

How do we reach a point where there are identical bases? We will break down the numbers correctly into prime factors.

For example:
8x=2x+28^x=2^{x+2}

We can break 88 down into factors and thus obtain that:
8=238=2^3
therefore:
(23)x=2x+2(2^3 )^x=2^{x+2}

According to the properties of powers we obtain that:
2x+2=23x2^{x+2}=2^{3x}

We have reached a point where the bases are identical! Now we can compare the powers and find the XX.

3x=x+23x=x+2
2x=22x=2
x=1x=1


Quadratic Equation Method and Placement of t - Example

To bring the equation to a state where there is one element, the same element squared, and a free number, we will need to use the properties of powers.
We will remember the following power rules:

(an)m=an×m(a^n )^m=a^{n\times m}

an+m=an×ama^{n+m}=a^n\times a^m

Let's take the following example:
42x+4x+1=04^{2x}+4^{x+1}=0
To bring the equation to a state where there is one element, the same element squared, and a free number, we will have to use power rules.
According to the power laws, we can say that:
42x=(4x)24^{2x}=(4^x)^2
In fact, we isolated 4x4^x and showed that it was quadratic.

We will also take 4x+14^{x+1}
According to the power laws, it can be said that:
4x+1=4×4x4^{x+1}=4\times 4^x

Now, let's rewrite the same equation with the data we received and we get:

(4x)2+4×4x=0(4^x )^2+4\times 4^x=0


Now, we place t t in place of the element with an unknown power.
Let's say that: t=4xt=4^x
Every time we use 4x4^x we replace it with, tt
Now, our equation will be simpler and will look like this:

t2+4×t=0t^2+4\times t=0
In fact, we have a simple quadratic equation!
Solve and find that:
t2+4t=0t^2+4t=0
t=4,t=0t=-4, t= 0


Pay attention!! This is not our final answer!
We have found the tt and not the unknown we are looking for XX.
To find xx we place in the equation t=4xt=4^x the results we obtained.
Attention! Both solutions were rejected because the power for a number that is not 00 cannot be 00.
The second solution is 4-4 disqualified because a power for a positive number always yields a positive result.

Therefore, there is no result!


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