How do you find the x intercept of $q(x) = ln ( x - 7 ) + 4$ ?

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Answer 1 · 2017-02-28T12:25:16

$x= e^-4+7 ~= 7.0183$

$q(x) = ln(x-7)+4$

The x intercept of $q(x)$ occurs where $q(x) = 0$

I.e. where: $ln(x-7)+4 = 0$

$ln(x-7) = -4$

Remember that $ln e =1$

$:. ln (x-7) = -4lne$

$ln (x-7) = ln e^-4$

Since: $ln a = ln b -> a=b$

$x-7 = e^-4$

$x= e^-4 +7 ~= 7.0183$

The x intercept can be seen on the graph of $g(x)$ below:

graph{ln(x-7) +4 [-1.625, 18.375, -2.6, 7.4]}

How do you find the x intercept of q(x) = ln ( x - 7 ) + 4q(x) = ln ( x - 7 ) + 4?