Question

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

(
1
,
0
)
,
(
1
16

,
2
)
,
and
(
4
,
−
1
)
(1,0),(116,2),and(4,-1)
.

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=√(116)\\\to b=√(4\cdot29)\to b=\sqrt4\cdot√(29)\to b=2√(29)\\\\For\ (4,\ -1)\to x=4,\ y=-1.\ Substitute:\\\\\log_b4=-1\to b^(-1)=4\to b=(1)/(4)`

Different values of b.

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

Maybe the second point is
`\left((1)/(16),\ 2\right)`

Substitute:

`\log_b(1)/(16)=2\to b^2=(1)/(16)\to b=\sqrt{(1)/(16)}\to b=(1)/(4)`

Then we have the answer:

`\boxed{y=\log_{(1)/(4)}x}`