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Dimas [21]
3 years ago
9

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

Mathematics
1 answer:
katovenus [111]3 years ago
6 0

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.

<h3>Answer: There is no logarithmic function whose graph goes through given points.</h3><h3 />

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}

<h3>Then we have the answer:</h3>

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

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