Exponential Equations are equations where at least part of the exponent is a variable. They are commonly expressed as such:
Exponential Equations consist of Exponential Expressions. Exponential functions and expressions can be expressed in different ways by changing the base. Changing the base of one or more exponential expressions is a useful technique for solving exponential equations.
Another way of solving exponential equations is by graphing using technology.
Solve the following exponential equation, \(2^{3x} = 8\).
First, we can express both terms as powers with base \(2\):
\(2^{3x} = 8\)
\(2^{3x} = 2^3\)
Next, since the bases of both terms are the same, we can equate the exponents to solve for \(x\):
\(3x = 3\)
\(x = 1\)
Therefore, we can determine that \(\boldsymbol{x = 1}\).
\(3^{x+2} = 81^x\)
First, we can express both terms as powers with base \(3\):
\(3^{x+2} = 81^x\)
\(3^{x+2} = (3^4)^x\)
Next, we can apply the Power of a Power Law to distribute the exponents on the right side:
Then, we can equate the exponents to solve for \(x\):
\(x+2 = 4x\)
\(3x = 2\)
\(x = \cfrac{2}{3}\)
Therefore, we can determine that \(\boldsymbol{x = \cfrac{2}{3}}\).
\(9^{x+1} = 27^{x-1}\)
First, we can express both terms as powers with base \(3\):
\(9^{x+1} = 27^{x-1}\)
\((3^2)^{x+1} = (3^3)^{x-1}\)
Next, we can apply the Power of a Power Law to distribute the exponents on both sides:
Then, we can equate the exponents to solve for \(x\):
\(2x + 2 = 3x - 3\)
\(x = 5\)
Therefore, we can determine that \(\boldsymbol{x = 5}\).
\(5^{2x+1} = 125\)
First, we can express both terms as powers with base \(5\):
\(5^{2x+1} = 125\)
\(5^{2x+1} = 5^3\)
Next, we can equate the exponents to solve for \(x\):
\(2x + 1 = 3\)
\(2x = 2\)
\(x = 1\)
Therefore, we can determine that \(\boldsymbol{x = 1}\).
An equation maintains balance when the Common Logarithm is applied to both sides. The Power Law of Logarithms is a useful tool for solving a variable that appears as part of an exponent. When a quadratic equation is obtained, methods such as factoring and applying the quadratic formula may be useful. Some algebraic methods such as solving exponential equations lead to Extraneous Roots, which are not valid solutions to the original equation.
For the equation, \(2^{t} = 5\), solve for \(t\). Round the answer to \(2\) decimal places.
First, we can take the common logarithm of both sides:
Next, we can apply the Power Law of Logarithms to bring down \(t\):
Finallly, we can solve for \(t\):
\(t = \cfrac{\log5}{\log2}\)
\(t \approx 2.322\)
Therefore, we can determine that \(\boldsymbol{t \approx 2.322}\).
\(5^{x+1} = 7^x\)
First, we can take the common logarithm of both sides:
Next, we can apply the Power Law of Logarithms to bring the exponents down:
Then, we can expand the expression by applying distributive property:
After, we can collect variable terms on one side of the equation and factor \(x\):
\(x\log7 - x\log5 = \log5\)
\(x(\log7 - \log5) = \log5\)
Finally, we can solve for \(x\):
Therefore, we can determine the solution in its exact form is \(\boldsymbol{x = \cfrac{\log5}{\log7 - \log5}}\).
\(2^{x - 4} = 9^{3x-1}\)
First, we can take the common logarithm of both sides:
Next, we can apply the power law of logarithms to bring down the exponents:
Then, we can expand the expression by applying distributive property:
After, we can collect variable terms on one side of the equation and factor \(x\):
\(3x\log9 - x\log2 = - 4\log2 + \log9\)
\(x(3\log9 - \log2) = - 4\log2 + \log9\)
Finally, we can solve for \(x\):
Therefore, we can determine the solution in its exact form is \(\boldsymbol{x = \cfrac{-4\log2 + \log9}{3\log9 - \log2}}\).