Match logarithmic functions and graphs Algebra I

Assume that:

y = log_a\left(x\right)

Since the point \left(9{,}2\right) is on the graph, we can plug in 9 for x and 2 for y and obtain:

2=log_a \left(9\right)

Therefore:

a^2 = 9

a=3

Thus:

y=log_3\left(x\right)

Assume that:

y = log_a\left(x\right)

Since the point \left(16{,}4\right) is on the graph, we can plug in 16 for x and 4 for y and obtain:

4=log_a \left(16\right)

Therefore:

a^4 = 16

a=2

Thus:

y=log_2\left(x\right)

Since the logarithm function is decreasing, we must have:

y = -log_a\left(x\right)

Since the point \left(5,-1\right) is on the graph, we can plug in 5 for x and -1 for y and obtain:

-1=-log_a \left(5\right)

Therefore:

a^1=5

We have:

y = \log_a\left(cx+d\right)

where a\ne 1 is a positive number and c,d are real numbers. Also, if we have a vertical asymptote at x=x_0, then we have cx_0+d=0.

There is a vertical asymptote at x=2 , Thus:

c\left(2\right)+d=0 \Rightarrow d=-2c

Therefore:

y=\log_a\left(cx-2c\right)

The only acceptable answer is:

y=\log_2\left(2x-4\right)

We have:

y = \log_a\left(x+d\right)

where a\ne 1 is a positive number and d is real a number.

The point \left(2{,}0\right) is on the graph, thus:

0=log_a \left(x+d\right) \Rightarrow 2+d =a^0 \Rightarrow 2+d=1 \Rightarrow d=-1

The point \left(4{,}1\right) is also on the graph. Thus:

1=\log_a\left(4-1\right) \Rightarrow a^1=3 \Rightarrow a=3

Therefore we have:

y=log_3\left(x-1\right)

We have:

y = \log_a\left(cx\right)

Where a\ne 1 is a positive number and c is real a number.

Since the point \left(\dfrac{1}{3},0\right) is on the graph. Thus:

0=\log_a\left(\dfrac{1}{3}c\right) \Rightarrow \dfrac{1}{3}c=a^0 \Rightarrow c=3

The point \left(3{,}1\right) is also on the graph:

1=\log_a \left(3x\right) \Rightarrow 3\times 3 =a^1 \Rightarrow a=9

Therefore:

y=\log_9\left(3x\right)

We have:

y = \log_a\left(cx+d\right)

where a\ne 1 is a positive number and c, d are real a number.s

The points \left(-1{,}0\right), \left(1{,}1\right) and \left(7{,}2\right) are on the graph, therefore:

\begin{cases} 0=\log_a\left(-c+d\right) \Rightarrow d-c=1 \cr \cr 1=\log_a\left(c+d\right) \Rightarrow c+d=a\end{cases}

Recall that if we have a vertical asymptote at x=x_0, then we have cx_0+d=0. Here, we have a vertical asymptote at x=-2. Therefore:

-2c+d=0 \Rightarrow d=2c

Thus:

\begin{cases} d-c=1 \Rightarrow 2c-c =1 \Rightarrow c=1 \, \text{and} \, d-1\cr \cr c+d=a \Rightarrow a=2 \end{cases}

y=\log_2\left(x-1\right)