Question

NO LINKS!! Please fill in the blanks. Part 6a.​

Answer 1

In a logarithmic equation, the variable is in the argument of the logarithm.

As with other types of equations, the goal in solving is to isolate the variable on one side of the equation.

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Example: Solve
`4\log_2(20-x) = 12` for x.

`\begin{array}c \cline{1-2} & \\\text{Isolate the log expression.} & \log_2(20-x) = 3\\ & \\\cline{1-2} & \\\text{Rewrite in exponential form.} & 2^3 = 20-x\\ & \\\cline{1-2} &\\\text{Simplify the exponent.} & 8 = 20-x\\ & \\\cline{1-2}\text{Solve for x.} & -12 = -x\\ & \\& 12 = x\\\cline{1-2}\end{array}`

Let's check the answer.

`4\log_2(20-x) = 12\\\\4\log_2(20-12) =12\\\\4\log_2(8) = 12\\\\4*(\log(8))/(\log(2)) = 12\\\\4*(0.90308998699194)/(0.30102999566399) \approx 12\\\\4*2.9999999999999 \approx 12\\\\11.9999999999996\approx 12`

We have rounding error. This is why we don't land exactly on 12, but we get really close.

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Or we could check the answer this way

`4\log_2(20-x) = 12\\\\\log_2(20-x) = 3\\\\\log_2(20-12) = 3\\\\\log_2(8) = 3\\\\8 = 2^3\\\\8 = 8 \ \ \ \checkmark\\\\`

The answer is confirmed.