2^k=n

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
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simplify solve for inverse tangent line

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solve for k, 2^k=n

Solution

k=ln(n)ln(2)

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2^k=n

Apply exponent rules

kln(2)=ln(n)

Solve kln(2)=ln(n): k=ln(n)ln(2)

k=ln(n)ln(2)

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solve for n, 2^k=n

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

2^k=n

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