Mendon, Utah

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Mendon is a city in Cache County, Utah, United States. The population was 898 at the 2000 census. It is included in the Logan, Utah-Idaho Metropolitan Statistical Area.



Mendon was settled in 1859.[3]


Mendon is located at 41°42′31″N 111°58′30″W / 41.70861°N 111.975°W / 41.70861; -111.975 (41.708738, -111.975127).[4]

According to the United States Census Bureau, the city has a total area of 1.2 square miles (3.2 km²), all of it land.


As of the census[2] of 2000, there were 898 people, 267 households, and 233 families residing in the city. The population density was 717.2 people per square mile (277.4/km²). There were 283 housing units at an average density of 226.0/sq mi (87.4/km²). The racial makeup of the city was 96.77% White, 1.22% Native American, 1.11% from other races, and 0.89% from two or more races. Hispanic or Latino of any race were 2.00% of the population.

There were 267 households out of which 50.2% had children under the age of 18 living with them, 80.1% were married couples living together, 4.9% had a female householder with no husband present, and 12.4% were non-families. 11.6% of all households were made up of individuals and 3.0% had someone living alone who was 65 years of age or older. The average household size was 3.36 and the average family size was 3.65.

In the city the population was spread out with 35.9% under the age of 18, 11.9% from 18 to 24, 23.7% from 25 to 44, 21.3% from 45 to 64, and 7.2% who were 65 years of age or older. The median age was 26 years. For every 100 females there were 102.3 males. For every 100 females age 18 and over, there were 100.0 males.

The median income for a household in the city was $46,563, and the median income for a family was $51,528. Males had a median income of $39,375 versus $20,795 for females. The per capita income for the city was $15,906. About 2.4% of families and 3.8% of the population were below the poverty line, including 3.1% of those under age 18 and 9.8% of those age 65 or over.

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