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

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>

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

6 0

3 years ago

Consider this equation (csc x+1)/cot x = cot x/(csc x +1) is it an identity?

levacccp [35]

Cross multiplying:-
cot^2 x = (csc x+ 1)(csc x + 1)

also cot^2 x = csc^2 x - 1 = (csc x + 1)(csc x - 1) (Known Identitiy)

so the equation is not an identity

7 0

3 years ago

BRAINLIST BRAINLISSSST

Alexxx [7]

Answer:

B is not true

Step-by-step explanation:

6 0

3 years ago

Please please please helppppppp the points graphed if you can’t see them are in the top right (5,5) bottom right (1,-2) bottom l

bonufazy [111]

Since there is one value of y for every value of x this is both a relation and a function

7 0

4 years ago