Graphs of Logarithmic Functions

If the following three curves intersect at one point, what is $a?$

$$\begin{array}{c}&y=\log_2 x, & y=\log_4 2x, &y=\log_8 ax\end{array}$$

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First we find the point at which $y=\log_2 x$ and $y=\log_4 2x$ intersect. Changing the base of the second equation gives

$$\begin{aligned} \log_4 2x&=\log_{2^2}2x\\ &=\frac{1}{2}\log_2 2x\\ &=\log_2 (2x)^{\frac{1}{2}}\\ &=\log_2 \sqrt{2x}. \end{aligned}$$

Then setting $\log_2 x=\log_2 \sqrt{2x}$ gives

$$x=\sqrt{2x} \implies x^2=2x,$$

which implies $x(x-2)=0\implies x=2$ since a logarithm function is defined over positive numbers. For $x=2,$ the value of $y=\log_2 x$ is $\log_2 2=1,$ implying that the first two curves intersect at the point $(2, 1).$

Now, for the third curve $\log_8 ax$ to pass through $(2, 1),$ it must be true that

$$\log_8 (a\cdot 2)=1,$$

which implies $2a=8.$ Therefore, our answer is $a=4.$ The graphs of the three functions would look like the figure below. $_\square$

log1

The curve $\log_{\frac{1}{3}} x$ is below the curve $\log_{\frac{1}{5}} x$ for $x>a.$ What is $a?$

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Since the logarithm of 1 always equals zero regardless of the base, we have

$$\log_{\frac{1}{3}}1=\log_{\frac{1}{5}}1=0,$$

so we know that the two curves meet at the point $(1,0).$ For any $x>1,$ we know that

$$\log_3 x>\log_5 x\implies \log_{\frac{1}{3}}x<\log_{\frac{1}{5}}x.$$

For any $x<1,$ we know that

$$\log_3 x<\log_5 x\implies \log_{\frac{1}{3}}x>\log_{\frac{1}{5}}x.$$

Therefore, the graphs of the two curves would look like the figure below:

log2

Since $\log_{\frac{1}{3}} x$ is below $\log_{\frac{1}{5}} x$ for $x>1,$ our answer is $a=1. \ _\square$

Which of the following is not correct about the curve $y=\log_3 (x+8)-5?$

$\text{(a) }$ The lowest value of $y$ on the curve is $y=-5.$
$\text{(b) }$ The curve overlaps $y=\log_3 x$ by a parallel translation.
$\text{(c) }$ The domain of the function $y=f(x)$ is $\{x \lvert x>-8\}.$
$\text{(d) }$ The function $y=f(x)$ is an increasing function.

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$\text{(a) }$ Since the range of the function $y=f(x)$ is $\{y \lvert -\infty < y < \infty\},$ this is not true.

$\text{(b) }$ Since the parallel translation of $y=\log_3 x$ by $-8$ in the positive direction of the $x$-axis and by $-5$ in the positive direction of the $y$-axis is $y=\log_3 (x+8)-5,$ this is true.

$\text{(c) }$ Since a logarithm function is defined over positive numbers, it must be true that $x+8>0,$ or $x>-8.$

$\text{(d) }$ $y=\log_3 x$ increases as $x$ increases, and so does $y=\log_3 (x+8)-5.$

log3

Therefore, our answer is $\text{(a)}.\ _\square$