How do you graph $g (x) = \frac{1}{2} {log}_{3} (x + 4) + 3$ $g(x)=1/2log_3 (x+4)+3$ ?

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Answer 1 · 2016-08-06T16:49:25

Use the inverse relation $x = 9^g/729-4,, g=0, +-9, +-9^2, +-9^3, ...$

Inversion:

$3^(2g-6)=x+4$

$x=9^g/9^3-4=9^(g-3)-4$

Now, use the Table for $g = 0, +-9, +-9^2, +-9^3, +-9^4$

$(x, g) = ( 9^(g-3)-4$ ):::... $(5, 4) (-2, 3) ( -35/9, 2) (-522/81, 1)$

$... to (-4, -oo)$

oo)#

$x = -4$ (downwards ) is asymptotic to the graph and $x >-4$ ....

How do you graph g(x)=1/2log_3 (x+4)+3g(x)=12log3(x+4)+3g(x)=1/2log_3 (x+4)+3?