What is exponential equation?
It is an equation whose exponent or part of the exponent is a variable.
Examples :
3^{x} = 27
5^{3x - 1} = 625
The following steps would be useful to solve exponential equations by rewriting the base.
Step 1 :
Using the rules of exponents, rewrite each side of the equation as a power with the same base.
Step 2 :
Once you get the same base on both sides in step 1, equate the exponents and solve for the variable.
a^{x} = a^{k}
x = k
Example 1 :
Solve for x :
3^{x} = 27
Solution :
3^{x} = 27
3^{x} = 3 ⋅ 3 ⋅ 3
3^{x} = 3^{3}
Equate the exponents.
x = 3
Example 2 :
Solve for y :
Solution :
2^{y - 5} = 2^{-4}
Equate the exponents.
y - 5 = -4
Add 5 to both sides.
y = 1
Example 3 :
Solve for x :
6 ⋅ 3^{x} = 54
Solution :
7 ⋅ 3^{x} = 63
Divide both sides by 7.
3^{x} = 9
3^{x} = 3^{2}
Equate the exponents.
x = 2
Example 4 :
Solve for z :
4^{z }= 7(2^{z}) + 18
Solution :
4^{z }= 7(2^{z}) + 18
(2^{2})^{z }= 7(2^{z}) + 18
(2^{z})^{2 }= 7(2^{z}) + 18
(2^{z})^{2 }- 7(2^{z}) - 18 = 0
Let a = 2^{z}.
a^{2 }- 7a - 18 = 0
(a - 9)(a + 2) = 0
a - 9 = 0 or a + 2 = 0
a - 9 = 0 a = 9 a^{ }= 3^{2} 3^{z }= 3^{2} z = 2 |
a + 2 = 0 a = -2 a = -2 3^{z }= -2 |
In 3^{z}, whatever real value (positive or negative or zero) we substitute for z, 3^{z} can never be negative. So we can ignore the equation 3^{z} = -2.
Therefore,
z = 2
Example 5 :
Solve for t :
5^{t }= (√25)^{-7 }⋅ (√5)^{-5})
Solution :
5^{t }= (√25)^{-7 }⋅ (√5)^{-5}
Equate the exponents.
t = -¹⁹⁄₂
Consider the following equation in exponential form.
b^{y} = a
The picture below illustrates how to convert the above equation from exponential to logarithmic form.
Still don't understand what is explained above, please watch the video below for step by step live explanation.
Example 6 :
Find the value of x, if 10^{x} = 100.
Solution :
10^{x} = 100
The above equation is in exponential form. Convert it to logarithmic form to solve for x.
x = log_{10}100
x = log_{10}100
x = log_{10}(10^{2})
x = 2log_{10}10
x = 2(1)
x = 2
Example 7 :
Find the value of x, if 3^{x} = ¹⁄₂₇.
Solution :
3^{x} = ¹⁄₂₇
The above equation is in exponential form. Convert it to logarithmic form to solve for x.
x = log_{3}(¹⁄₂₇)
x = log_{3}1 - log_{3}27
x = 0 - log_{3}(3^{3})
x = -3log_{3}3
x = -3(1)
x = -3
Example 8 :
Find the value of x, if 5^{3x - 1} = ¹⁄₆₂₅.
Solution :
5^{3x - 1} = ¹⁄₆₂₅
The above equation is in exponential form. Convert it to logarithmic form to solve for x.
3x - 1 = log_{5}(¹⁄₆₂₅)
3x - 1 = log_{5}1 - log_{5}625
3x - 1 = 0 - log_{5}(5^{4})
3x - 1 = -4log_{5}5
3x - 1 = -4(1)
3x - 1 = -4
Add 1 to both sides.
3x = -3
Divide both sides by 3.
x = -1
When we solve exponential equations, we will use the rules of exponents and rewrite each side of the equation as a power with the same base.
Incase you are not able to rewrite each side of the equation as a power with the same base, take logarithm on both sides of the equation and solve for x.
Example 9 :
Solve for x :
2^{x} - 1 = 5
Solution :
2^{x} - 1 = 5
Add 1 to both sides.
2^{x} = 5
On the left side of the equation above, the base is 2. On the right side, 6 is not a power of 2. So, we can not get the same base on both sides.
Take logarithm on both sides and solve for x.
log(2^{x}) = log6
Using power rule of logarithm,
xlog2 = log6
Divide both sides by log2.
Example 10 :
Solve for x :
5^{x - 1 }- 2^{x} = 0
Solution :
5^{x - 1 }- 2^{x} = 0
Add 2^{x} to both sides.
5^{x - 1 }= 2^{x}
In the equation above, the base on the right side and the base left side are not same. And also, we can not make the base same on both sides using the rules of exponents.
Take logarithm on both sides and solve for x.
log(5^{x - 1}) = log(2^{x})
Using power rule of logarithm,
(x - 1)log5^{ }= xlog2
Using distributive property,
xlog5 - log5^{ }= xlog2
Subtract xlog2 from both sides.
xlog5 - log5^{ }- xlog2 = 0
Add log5 to both sides.
xlog5^{ }- xlog2 = log5
Factor.
x(log5^{ }- log)2 = log5
Divide both sides by (log5 - log2).
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