Solving Exponential Equations with the Same Base or Like Base

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 base.


Key Steps in Solving Exponential Equations without Logarithms

Make the base on both sides of the equation the SAME

so that if [latex]\large{b^{\color{blue}M}} = {b^{\color{red}N}}[/latex]

then [latex]{\color{blue}M} = {\color{red}N}[/latex]

  • In other words, if you can express the exponential equations to have the same base on both sides, then it is okay to set their powers or exponents equal to each other.

You should also remember the properties of exponents in order to be successful in solving exponential equations.


Basic Properties of Exponents

1) Zero Property

Any nonzero number to the power of zero is just 1. In equation it looks like this, b^0 = 1 where b ≠ 0.

2) Negative Exponent Property

For any nonzero number "b", and any integer n, 1 divided by b raised to the power of negative n is the reciprocal of b raised to the power of n. In symbols, b^-n = 1/(b^n) or 1/(b^-n) = b^n, however, b ≠ 0.

3) Product Rule

For any nonzero number "b" and any integers m and n, the product of b to the power of m , and b to the power of n is equal to b raised to the power of the sum of m and n. In symbols, (b^m) * (b^n) = b^(m+n) where b ≠ 0.

4) Quotient Rule

For any nonzero real number "b" and any integers m and n, the quotient of b to the power of m, and b to the power of n is equal to b raised to the power of the difference of m and n. In math form, we can express this particular exponent rule, to be specific quotient rule, as b^m/b^n = b^(m-n) but b≠0.

5) Power to a Power Rule

For any nonzero real number b and any integers m and n, the quantity of b to the power of m raised to the power of n is simply b to the power of the product of m and n. We can express this exponent rule as (b^m)^n = b^(mn).

Let’s take a look at some examples!


Examples of How to Solve Exponential Equations without Logarithms

Example 1: Solve the exponential equation below using the Basic Properties of Exponents.

solve the following exponential equation using exponent rules: 5^(3x) = 1/125

Solution:

  • Given
it is given that 5 raised to the power of 3x is equal to 1 over 125, or simply, 5^(3x)=(1)/(125)
  • Express the denominator of the right side with a base of [latex]5[/latex]. We have [latex]125 = {5^3}[/latex].

Apply the Negative Exponent Property.

5^(3x)=1/(5^3) since 5^3 = 125
  • At this point, the bases are the same therefore set the powers equal to each other.
applying the negative rule of exponent we get 5^(3x) = 5^(-3)
  • This is just a simple one-step linear equation.
because the base is the same on both sides of the equation, we can set their exponent equal to each other. therefore, 3x = -3.
  • To solve for [latex]x[/latex], divide both sides by [latex]3[/latex]. That’s it!
3x = -3 ==> (3x)/3 = (-3)/(3) ==> x = -1

The final answer here is [latex]x = – 1[/latex].


Example 2: Solve the exponential equation below using the Basic Properties of Exponents.

use the exponent rules to solve for the exponential equation  * 8 = 256^x

Solution:

  • Given
the quantity 2 to the power of 7x times 8 equals 256 to the power of x
  • Express all numbers with the base of [latex]2[/latex]. So we have: [latex]8 = {2^3}[/latex] and [latex]256 = {2^8}[/latex].

Apply the Product Rule on the left, while using the Power to a Power Rule on the right side.

the quantity 2 to the power of 7x multiplies to the 2 raised to the third power is equal to the quantity 2 to the eight power raised to the power of x. writing this in math format, we have  * 2^3 = ^x
  • 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.
2 to the power of sum of 7x and 3 is equal to 2 raised to the power of 8x. therefore we can write this as 2^(7x+3) = 2^(8x)
  • Solve the simple linear equation.
both sides of the equation has the same base of 2, that implies that we can set their corresponding exponents equal to each other. thus, 7x+3 = 8x.
  • Subtract both sides by [latex]7x[/latex] to isolate [latex]x[/latex]. Done!
7x-7x+3=8x-7x results to 3=x or x=3

The final answer is [latex]x=3[/latex].


Example 3: Solve the exponential equation below using the Basic Properties of Exponents.

/[16^(2x+1)] = 1

Solution:

  • Given
the given exponential equation to solve is 64 to the power of (2x) divided by 16 to the power of (2x+3) is equal to 1
  • Express each number with a base of [latex]2[/latex]. In doing so…

[latex]64 = {2^6}[/latex] and [latex]16 = {2^4}[/latex]

^(2x) / ^(2x+3)] =1
  • 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.

(2^12x)/(2^8x+12)=1
  • Apply the Quotient Rule.

Subtract the top exponent by the bottom exponent.

2^[12x-(8x+12)]=1
  • This is how it looks after subtracting the exponents.
2^(4x-12)=1

Now, looking at the right side, can we express [latex]1[/latex] as an exponential number with base [latex]2[/latex]?

The answer is yes! We can write it as [latex]1 = {2^0}[/latex] using the Zero Property of Exponent.

  • Now we have the set-up that we want – having the same bases on both sides.
2^(4x-12)=2^0
  • Set the exponent of the left-hand side of the equation equal to the exponent of the right-hand side, then solve the equation for the variable [latex]x[/latex].
4x-12=0
  • To solve the equation, start by adding both sides by [latex]12[/latex] to move the constant to the right side.
  • Finally, divide both sides by [latex]4[/latex] to get the value of [latex]x[/latex].
4x-12+12 = 0+12 results to x=3

The final answer is [latex]x = 3[/latex].


Example 4: Solve the exponential equation below.

[(1/36)^3-x][(1/6)^x]=216

Solution:

  • Start by writing the equation to solve.
[(1/36)^3-x][(1/6)^x]=216
  • Express each fraction as an exponential number with a base of [latex]6[/latex].

[latex]36 = {6^2}[/latex]

[latex]6 = {6^1}[/latex]

[latex]216 = {6^3}[/latex]

[latex]\Large{\left( {{1 \over {{6^2}}}} \right)^{3 – x}}{\left( {{1 \over {{6^1}}}} \right)^x} = {6^3}[/latex]

  • Apply the Negative Exponent Property on the left side of the equation.
[(6^-2)^3-x]{[(6)^-1]^x}=6^3
  • Multiply the inner exponents to outer exponents using the Power to a Power Rule.
[(6)^-6+2x][(6)^-x]=6^3
  • Since they have a common base, add the exponents using the Product Rule.
[(6)^-6+2x][^-x]=6^3
  • It’s obvious that by having a single and the same base on both sides, we can now set each power equal to each other.
(6)^(-6+x)=6^3
  • Solve the linear equation by adding both sides by [latex]6[/latex] to get [latex]x = 9[/latex].
-6x+x = 3 results to x=9

And so the solution is [latex]x = 9[/latex].


Example 5: Solve the exponential equation below using the Basic Properties of Exponents.

9 to the power of (-3x+2) times 3 to the power of (-x) is equal to 27 to the power of (-2x-1). in short form, we write this exponential equation as 9^(-3x+2) * 3^(-x) = 27^(-2x-1).

Solution:

  • Given
9 raised to the power of the sum of 2 and -3x multiplied to 3 to the power of -x is equal to 27 to the power of the sum of -2x and -1
  • Use [latex]3[/latex] as the common base.

[latex]9={3^2}[/latex] and [latex]27={3^3}[/latex]

  • Multiply the inner and outer exponents by applying the Power to a Power Rule.
the quantity 3 to the power of 2 to the power of -3x+2 multiplied to the quantity 3 to the power of -x is equal to the quantity 3 to the power of 3 to the power of -2x-1
  • At this point, we can add the exponents on the left side of the equation because they now have common bases.
3^(-6x+4) times 3^(-x) = 3^(-6x-3)
  • Apply the Product Rule by adding the exponents when bases are equal.
3^ = 3^(-6x-3)
  • Clearly, we can set the powers of both sides of the equation equal to each other.
3^(-7x+4) = 3^(-6x-3)
  • This results in a simple multi-step equation.
-7x+4=-6x-3
  • So we add [latex]6x[/latex] first on both sides. Then, subtract by [latex]4[/latex]. And finally, divide by [latex] – 1[/latex] to fully isolate [latex]x[/latex] by itself!
here are the step-by-step solutions to solving the linear equation: -7x+6x+4=-6x+6x-3 ==>-x+4=-3==> -x+4-4=-3-4 ==> -x=-7 ==> x=7

The answer is [latex]x=7[/latex]. Easy!


Example 6: Solve the exponential equation below using the Basic Properties of Exponents.

this is the description of the exponential equation not requiring logarithms to solve: the quantity 32 to the power of x^2 + 1 multiplied to the quantity 8 to the power of x-5 is equal to 16 to the power of 2x

Solution:

  • Given
  = 16^(2x)
  • Express each number with a base of [latex]2[/latex].

Next, multiply the inner exponents to outer exponents using the Power to a Power Rule.

the quantity 2 to the power of 5 to the power of x^2+1 multiplied to the quantity 2 to the power of 3 to the power of x-5 is equal to the quantity 2 to the power of 4 to the power of 2x
  • To generate a single base on the left side, use the Product Rule – copy the common base [latex]2[/latex] and add the exponents.
2^(5x^2+5) * 2^(3x-15) = 2^(8x)
  • This is when we apply the Product Rule.
this is an exponential equation with a common base of 2 which can be solved without the use of logarithms. we just need to set the exponents equal to each other since the base is the same then solve the linear equation: 2^ = 2^8x
  • After the addition of exponents, we have single bases on each side.

It’s time to set the powers equal to each other.

2 raised to the power of the 5x^2+3x-10 is equal to 2 raised to the power of 8x
  • 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.

5x^2+3x-10=8x
  • Solve the quadratic equation using the factoring method. Factor out [latex]5[/latex] in the trinomial then factor out the simple trinomial as a product of two binomials.
set 5x^2-5x-10 to zero. factor out 5 to get 5(x^2-x-2)=0. then factor out the binomial inside the parenthesis to obtain the quadratic equation 5(x-2)(x+1)=0
  • Using the Zero Property, we get these values for [latex]x[/latex].
x=2 or x=-1

The correct answers are [latex]x = 2[/latex] and [latex]x = – 1[/latex].


Example 7: Solve the exponential equation below using the Basic Properties of Exponents.

(1/49)^(-6x) times 49 is equal to the square root of 7

Solution:

  • Given
 (49) = sqrt (7)
  • Express each number as the exponential number with a base of [latex]7[/latex].
 (7^2) = sqrt(7)
  • Apply the Negative Exponent Property on the left side.

Also, the square root symbol can be rewritten as the exponent of [latex]\large{1 \over 2}[/latex].

the quantity 7 raised to the power of -2 to the power of -6x multiplied to 7 to the 2nd power is equal to 7 to the power of one-half
  • Apply the Power to a Power Rule on the left side.
 *( 7^2) = 7^(1/2)
  • Express the left side with a single base using the Product Rule by copying the common base and adding the exponents.
7^(12x+2) = 7^(1/2)
  • We can now set the powers equal to each other, then solve.
12x + 2 = 1/2
  • To solve for [latex]x[/latex], subtract both sides by [latex]2[/latex].
after setting the exponents equal to each other, we have 12x+2 = 1/2. Now subtract both sides by 2 to get 12x+2-2 = 1/2 -2.
  • Simplify
12x = -3/2
  • To finish this off, divide both sides by [latex]12[/latex].
x=(-1)/8

The final solution is [latex]x = – {\large{1 \over 8}}[/latex].


Example 8: Solve the exponential equation below using the Basic Properties of Exponents.

the quantity 125 raised to the power of (-4x+5) divided by the quantity 25 raised to the power of x^2-5x+1 equals 5

Solution:

  • Given
(125)^(-4x+5)/(25)^(x^2-5x+1) = 5
  • Express the numbers using the base [latex]5[/latex].

Next, multiply the inner and outer exponents using the Power to a Power Rule.

(5^3)^(-4x+5)/(5^2)^(x^2-5x+1) = 5
  • It looks like we can use the Quotient Rule because we have the same bases on the numerator and denominator.
the quantity 5 to the power of (-12x+15) divided by the quantity 5 to the power of (2x^2-10x+2) is equal to 5
  • Subtract the exponent on the numerator by the exponent in the denominator.
5 raised to the power  is equal to the constant 5
  • Simplify

It’s okay now to set the “powers” or exponents equal to each other and then solve the quadratic equation.

the exponential equation has a common base of 5. therefore, we can set their exponents equal to one another then solve. 5^(-2x^2-2x+13) = 5^1.
  • Solve the quadratic equation by factoring the trinomial into two binomials. Then set each binomial equal to [latex]0[/latex] to solve for [latex]x[/latex].
-2(x+3)(x-2)=0
  • Using the Zero Product Property, we obtain these values of [latex]x[/latex].
x is equal to -3 or x is equal to 2

The final answers are [latex]x = – 3[/latex] and [latex]x = 2[/latex].


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Solving Exponential Equations using Logarithms