Question

Find the logarithmic function y = logbx that passes through the points

Answer 1

$b\in(0,\ 1)\ \cup\ (1,\ \infty)\\x > 0\\y\in\mathbb{R}\\--------------------------\\\\y=\log_bx\\\\For\ (1,\ 0)\to x=1,\ y=0.\ Substitute:\\\\\log_b1=0\to b^0=1\to b\in(0,\ 1)\ \cup\ (1,\ \infty)\\\\For\ (116,\ 2)\to x=116,\ y=2.\ Substitute:\\\\\log_b116=2\to b^2=116\to b=\sqrt{116}\\\to b=\sqrt{4\cdot29}\to b=\sqrt4\cdot\sqrt{29}\to b=2\sqrt{29}\\\\For\ (4,\ -1)\to x=4,\ y=-1.\ Substitute:\\\\\log_b4=-1\to b^{-1}=4\to b=\dfrac{1}{4}$

Different values of b.

Answer: There is no logarithmic function whose graph goes through given points.

Maybe the second point is $\left(\dfrac{1}{16},\ 2\right)$

Substitute:

$\log_b\dfrac{1}{16}=2\to b^2=\dfrac{1}{16}\to b=\sqrt{\dfrac{1}{16}}\to b=\dfrac{1}{4}$

Then we have the answer:

$\boxed{y=\log_{\frac{1}{4}}x}$