How do you graph $y = 5 ln (3 x)$ $y=5ln(3x)$ ?

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Answer 1 · 2016-08-24T02:06:21

y is defined for $x > 0$ $x>0$ ; y has a zero at $x = \frac{1}{3}$ $x=1/3$ ;
$y \to - \infty$ $y -> -oo$ as $x \to 0$ $x-> 0$ (See graphs below)

$y = 5 ln (3 x)$ $y=5ln(3x)$

Since $ln a$ $lna$ is defined for $a > 0 \to y$ $a>0 -> y$ is defined for $3 x > 0$ $3x>0$
Hence for $x > 0$ $x>0$

Since $ln a = 0$ $lna=0$ for $a = 1 \to y = 0$ $a=1 -> y =0$ for $3 x = 1$ $3x =1$
Hence y has a zero at $x = \frac{1}{3}$ $x=1/3$

For the purposes of graphing function $y$ $y$ the $5$ $5$ has the effect of amplifying the magnitude of $ln 3 x$ $ln3x$ five times.

graph{5ln(3x) [-46.2, 46.2, -23.1, 23.15]}

In the region closer to $(\frac{1}{3}, 0)$ $(1/3, 0)$

graph{5ln(3x) [-1.915, 1.927, -0.941, 0.983]}

How do you graph y=5ln(3x)y=5ln(3x)?