| Solving Exponential Equations without Logarithms |
An exponential equation involves an unknown variable in the exponent. In this lesson, we will focus on the exponential equations that do not require the use of logarithm. In algebra, this topic is also known as solving exponential equations with the same base. Why? The reason is that we can solve the equation by forcing both sides of the exponential equation to have the same or equal base.
There are eight (8) worked examples in this lesson.
Key Steps
Make the base in both sides of the equation the SAME
so that if bM = bN
then M = N
- In other words, if you can express the exponential equations to have the same base in both sides then it's OKAY to set their powers or exponents equal to each other.
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You should also remember the properties of exponents in order to be successful in solving exponential equations. Here they are...
| Direction: Solve each exponential equation using the Basic Properties of Exponents. |
Example 1: Solve the exponential equation .
Solution:
The final answer here is x = −1.
Example 2: Solve the exponential equation .
Solution:
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Given |
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Express all numbers with the base of 2. So we have: 8 = 23 and 256 = 28.
Apply the Product Rule on the left, while using the Power to a Power Rule on the right side. |
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Here we are ready to set the powers equal to each other since we are able to create single bases that are the same on both sides. |
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Solve the simple linear equation. |
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Subtract both sides by 7x to isolate x. Done! |
The final answer is x = 3.
Example 3: Solve the exponential equation .
Solution:
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Given |
![[(2^6)^2x]/[(2^4)^(2x+3)]]=1](images/expoeq_ex3_step1_1.gif) |
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Express each number with a base of 2. In doing so...
64 = 26
and 16 = 24
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![[2^12x/2^(8x+12]=1](images/expoeq_ex3_step1.gif) |
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Applying Power to a Power Rule.
In other words, multiply the inner exponent to the outer exponent. Do it for both the numerator and denominator. |
![2^[12-(8x+12)]=1](images/expoeq_ex3_step2.gif) |
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Apply the Quotient Rule.
Subtract the top exponent by the bottom exponent. |
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This is how it looks after subtracting the exponents.
Now, looking at the right side, can we express 1 as an exponential number with base 2?
The answer is yes! 1 = 20 using the Zero Property of Exponent. |
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Now we have the set-up that we want - having the same bases in both sides. |
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Set the powers equal to each other, then solve the equation. |
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To solve the equation, start by adding both sides by 12 to move the constant to the right side.
Finally, divide both sides by 4 to get the value of x. |
Okay, so we find the answer to be x = 3.
Example 4: Solve the exponential equation .
Solution:
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Given |
![[1/6^2]^(3-x)*[1/6^1]^x=6^3](images/expoeq_ex4_step1.gif) |
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Express each number as an exponential number with base 6.
36 = 62
6 = 61
216 = 63
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![[[6^-2)^(3-x)*(6^-1)^x]=6^2](images/expoeq_ex4_step2.gif) |
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Apply the Negative Exponent Property on the left side of the equation. |
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Multiply the inner exponents to outer exponents using the Power to a Power Rule. |
![6^[(-6+2x)+(-x)]=6^3](images/expoeq_ex4_step4.gif) |
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Since they have a common base, add the exponents using the Product Rule. |
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It's obvious that by having a single and equal bases on both sides, we can now set each powers equal to each other. |
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Solve the linear equation by adding both sides by 6 to get x = 9. |
And so the solution is x = 9.
Example 5: Solve the exponential equation .
Solution:
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Given |
![[3^2]^(-3x+2)*3^-x=[3^3]^(-2x-1)](images/expoeq_ex5_step1.gif) |
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Use 3 as the common base.
9 = 32 and 27 = 33
Multiply the inner and outer exponents by applying the Power to a Power Rule.
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At this point, we can add the exponents on the left side of the equation because they now have common bases. |
![3^[(-6x+4)+(-x)]=3^(-6x-3)](images/expoeq_ex5_step3.gif) |
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Apply the Product Rule by adding the exponents when bases are equal. |
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Clearly, we can set the powers of both sides of the equation equal to each other. |
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This results to a simple multistep equation. |
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So we add 6x first on both sides. Then, subtract by 4. And finally, divide by -1 to fully isolate x by itself! |
The answer is x = 7. Easy!
Example 6: Solve the exponential equation .
Solution:
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Given |
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Express each number with a base of 2.
Next, multiply the inner exponents to outer exponents using the Power to a Power Rule. |
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To generate a single base on the left side, use the Product Rule - copy the common base 2 and add the exponents. |
![2^[(5x^2+5)+(3x-15)]=2^8x](images/expoeq_ex6_step3.gif) |
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This is when we apply the Product Rule. |
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After the addition of exponents, we have single bases on each side.
It's time to set the powers equal to each other. |
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After equating the powers, we arrive at this quadratic equation.
We need to move all terms on one side while forcing the opposite side equal to zero. |
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Solve the quadratic equation using factoring method. |
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Using the Zero Property, we get these values for x. |
The correct answers are x = 2 and x = −1.
Example 7: Solve the exponential equation .
Solution:
The final answer is x = -1/8.
Example 8: Solve the exponential equation .
Solution:
![[(125)^(-4x+5)]/[(25)^(x^2-5x+1)]=5](data:image/png;base64,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) |
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Given |
![[5^3]^(-4x+5)/[5^2]^(x^2-5x+1)=5](images/expoeq_ex8_step1.gif) |
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Express the numbers using the base 5.
Next, multiply the inner and outer exponents using the Power to a Power Rule.
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![[5]^(-12x+15)/[5]^(2x^2-10x+2)=5](images/expoeq_ex8_step2.gif) |
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It looks like that we can use the Quotient Rule because we have the same bases on the numerator and denominator.
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Subtract the exponent on the numerator by the one in the denominator. |
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Simplify
It's okay now to set the powers equal to each other, and then solve the equation. |
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Solve the quadratic equation using factoring method. |
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Using the Zero Product Property, we obtain these values of x. |
The final answers are x = −3 and x = 2.
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