Logarithmic Equations are functions that include a logarithm of any other positive base in place of regular terms such as \(x\). They are commonly expressed as such:
It is possible to solve an equation involving logarithms by expressing both sides as a logarithm of the same base: if \(a=b\), then \(\loga= \log b\) and vice versa.
We can find the roots of a logarithmic function either algebraically or graphing using technology.
Solve the logarithmic equation, \(\log(x + 4) = 1\).
First, we can rewrite the equation in exponential form, using base \(10\):
\(x + 4 = 10^1\)
\(x + 4 = 10\)
We can now solve for \(x\):
Therefore, we can determine that \(\boldsymbol{x = 6}\).
\(3\log(t - 52) = 9\)
First, we can divide both sides by \(3\):
First, we can rewrite the equation in exponential form, using base \(10\):
\(t - 52 = 10^3\)
\(t - 52 = 1000\)
We can now solve for \(t\):
Therefore, we can determine that \(\boldsymbol{t = 1052}\).
\(\log_4(s + 3) + 2 = 0\)
First, we can move \(2\) onto the opposite side:
Next, we can rewrite the expression in exponential form, using base \(6\):
\(s + 3 = 4^{-2}\)
\(s + 3 = \cfrac{1}{16}\)
We can now solve for \(s\):
Therefore, we can determine that \(\boldsymbol{s = - \cfrac{47}{16}}\).
When a quadratic equation is obtained, methods such as factoring or applying the quadratic formula may be useful.
Some algebraic methods of solving logarithmic equations lead to extraneous roots, which are not valid solutions to the original equation.
Solve the logarithmic equation, \(\log_5(x+6) + \log_5x = 2\). Identify and reject any extraneous roots.
First, we can apply the product law of logarithms:
\(\log_5(x+6) + \log_5x = 2\)
\(\log_5[(x+6)x] = 2\)
Next, we can expand the product of binomials on the left side. We can also rewrite the equation in exponential form, using base \(4\):
\(x^2 + 6x = 5^2\)
\(x^2 + 6x = 25\)
Then, we can move all terms onto the same side to express it in Standard Form:
Finally, we can find the roots by using the quadratic formula. From the formula above, we can determine that \(\textcolor{red}{a=1}\), \(\textcolor{green}{b=6}\), and \(\textcolor{blue}{c=-25}\):
Since this equation gives 2 solutions, we can explore each case.
For the first case, we can add the terms:
\(x_1 = -3 + \sqrt{34}\)
\(x_1 \approx 2.831\)
For the second case, we can subtract the terms:
\(x_2 = -3 - \sqrt{34}\)
\(x_2 \approx -8.831\)
We can reject \(x \approx -8.831\) as an extraneous root. Both \(\log_5(x+6)\) and \(\log x\) are undefined for this value since the logarithm of a negative number is undefined. Therefore, the only solution is \(\boldsymbol{x \approx 2.831}\).
\(\log5x^2 - \log4 = \log x\)
First, we can apply the Quotient Law of Logarithms:
Next, we can drop the log on both sides:
Then, we can multiply both sides by \(4\) and bring all terms to one sides:
\(5x^2 = 4x\)
\(5x^2 - 4x = 0\)
Finally, we can factor the expression to determine its roots:
\(x(5x - 4) = 0\)
\(x_1 = 0\)
\(x_2 = \cfrac{4}{5}\)
We can reject \(x=0\) as an extraneous root. Both \(\log 5x^2\) and \(\log x\) are undefined for this value since the logarithm of a negative number is undefined. Therefore, the only solution is \(\boldsymbol{x = \cfrac{4}{5}}\).
\(\log_3(x + 1) - \log_3(x+ 2) = \log_3(2x - 1) - \log(5x + 4)\)
First, we can apply the Quotient Law of Logarithms:
Next, we can drop the logs and set the arguments equal to each other:
Since this equation is rational, we can simplify it by cross-multiplying. We can then simplify the expression by using distributive property:
\((x+3)(2x - 2) = (x + 5)(x)\)
\(2x^2 + 4x - 6 = x^2 + 5x\)
Then, we can move all terms onto one side and simplify by collecting like terms:
Finally, we can find the factors of this expression, which represent the potential solutions:
\((x - 3)(x + 2) = 0\)
\(x_1 = 3\)
\(x_2 = -2\)
We can reject \(x=-2\) as an extraneous root. Both \(\log_3(x + 1)\) and \(\log_3(x+ 2)\) are undefined for this value since the logarithm of a negative number is undefined. Therefore, the only solution is \(\boldsymbol{x = 3}\).