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Task 5. Using the Laws of Logarithms to Solve Logarithmic Equations. We will use the Laws of Logarithms to simplify the equation as far as possible and then use the definition to solve for the unknown argument. Here is an example: Solve the equation $log_2(x) - log_2(x - 3) = 1$ Simplify the equation: $log_2(frac{x}{x - 3}) = 1$ Use the definition, the inverse function, to solve for the argument: $frac{x}{x - 3} = 2^1$ Continue to solve the algebraic equation: $x = 2x - 6$ $x = 6$ Check the solution to see that it is in the domain: $log_2(6) - log_2(6 - 3) = 1$ $log_2(6) - log_2(3) = 1$ You can proceed to check this further using a calculator or a hand calculation. But it should work since we can see that the terms $log_2(6)$ and $log_2(3)$ are defined, that is we know that both 6 and 3 are in the domain of the logarithm. Solve the following equations. Be sure to check the potential solution(s) in the original equation, to see whether it (they) are in the domain. (a) $log_2(x^2 - x - 2) = 2$ (b) $2log(x) - log(2) - log(3x - 4) = 0$ (c) $log(x) + log(x - 3) = 1$

          Task 5. Using the Laws of Logarithms to Solve Logarithmic Equations.
We will use the Laws of Logarithms to simplify the equation as far as possible and then use the
definition to solve for the unknown argument. Here is an example: Solve the equation
$log_2(x) - log_2(x - 3) = 1$
Simplify the equation:
$log_2(frac{x}{x - 3}) = 1$
Use the definition, the inverse function, to solve for the argument:
$frac{x}{x - 3} = 2^1$
Continue to solve the algebraic equation:
$x = 2x - 6$
$x = 6$
Check the solution to see that it is in the domain:
$log_2(6) - log_2(6 - 3) = 1$
$log_2(6) - log_2(3) = 1$
You can proceed to check this further using a calculator or a hand calculation. But it should work since
we can see that the terms $log_2(6)$ and $log_2(3)$ are defined, that is we know that both 6 and 3 are in the
domain of the logarithm.
Solve the following equations. Be sure to check the potential solution(s) in the original
equation, to see whether it (they) are in the domain.
(a) $log_2(x^2 - x - 2) = 2$
(b) $2log(x) - log(2) - log(3x - 4) = 0$
(c) $log(x) + log(x - 3) = 1$

$Task 5. Using the Laws of Logarithms to Solve Logarithmic Equations. We will use the Laws of Logarithms to simplify the equation as far as possible and then use the definition to solve for the unknown argument. Here is an example: Solve the equation log2(x) - log2(x - 3) = 1 Simplify the equation: log2(fracxx - 3) = 1 Use the definition, the inverse function, to solve for the argument: fracxx - 3 = 2^1 Continue to solve the algebraic equation: x = 2x - 6 x = 6 Check the solution to see that it is in the domain: log2(6) - log2(6 - 3) = 1 log2(6) - log2(3) = 1 You can proceed to check this further using a calculator or a hand calculation. But it should work since we can see that the terms log2(6) and log2(3) are defined, that is we know that both 6 and 3 are in the domain of the logarithm. Solve the following equations. Be sure to check the potential solution(s) in the original equation, to see whether it (they) are in the domain. (a) log2(x^2 - x - 2) = 2 (b) 2log(x) - log(2) - log(3x - 4) = 0 (c) log(x) + log(x - 3) = 1$

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