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# How do you find the domain, x intercept and vertical asymptotes of y=-log_3x+2?

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VNVDVI Share
Apr 8, 2018

Domain: $\left(0 , \infty\right)$ $x -$intercept: $\left(9 , 0\right)$ Vertical asymptote: $x = 0$

#### Explanation:

Recalling that the logarithmic function (in this case ${\log}_{3} x$) does not exist for negative $x$ or $x = 0$, the domain is all $x > 0 ,$ or $\left(0 , \infty\right)$

The vertical asymptote will be at $x = 0$, as the function is never truly defined for this value, but gets infinitely close to it.

The $x -$intercept can be found by setting the function equal to zero and solving for $x .$

$- {\log}_{3} x + 2 = 0$

${\log}_{3} x = 2$

Recalling that ${\log}_{a} b = \ln \frac{b}{\ln} \left(a\right)$, we rewrite as

$\ln \frac{x}{\ln} \left(3\right) = 2$

$\ln \left(x\right) = 2 \ln 3$

$2 \ln \left(3\right) = \ln \left({3}^{2}\right) = \ln \left(9\right)$

So,

$\ln \left(x\right) = \ln \left(9\right)$

$x = 9$

The $x -$intercept is $\left(9 , 0\right)$

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