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**Unformatted text preview: ** x) + 3, we gather the logarithms to one side and get
x+3
log2 (x + 3) − log2 (6 − x) = 3, and then use the Quotient Rule to obtain log2 6−x = 3.
Rewriting this as an exponential equation gives 23 = x+3
6−x . This reduces to the linear equation 8(6 − x) = x + 3, which gives us x = 5. When we graph f (x) =
we ﬁnd they intersect at x = 5. ln(x+3)
ln(2) and g (x) = ln(6−x)
ln(2) + 3, 370 Exponential and Logarithmic Functions y = f (x) = log7 (1 − 2x) and
y = g (x) = 1 − log7 (3 − x) y = f (x) = log2 (x + 3) and
y = g (x) = log2 (6 − x) + 3 6. Starting with 1 + 2 log4 (x + 1) = 2 log2 (x), we gather the logs to one side to get the equation
1 = 2 log2 (x) − 2 log4 (x + 1). Before we can combine the logarithms, however, we need a
common base. Since 4 is a power of 2, we use change of base to convert log4 (x+1) = log2 (x+1) =
log (4)
1
2 log2 (x + 1). Hence, our original equation becomes 1 = 2 log2 (x) − 2 1
2 2 log2 (x + 1) or 1 = 2 log2 (x) − log2 (x + 1). Using the...

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