ln(2x-3)+ln(x-1)>0

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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ln(2x−3)+ln(x−1)>0

Solution

x>2

Interval Notation

(2, ∞ )

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Solution steps

One step at a time

ln(2x−3)+ln(x−1)>0

Apply log rules

(2x−3)(x−1)>1

(2x−3)(x−1)>1 : x<12 or x>2

x<12 or x>2

Find positive values for logs: x>32

Combine the intervals

(x<12 or x>2) and x>32

Merge Overlapping Intervals

x>2

Number Line

x>2

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

ln(2x-3)+ln(x-1)>0

Solution

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