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Fråga 10 av 10 Gunilla påstår att \( \lg x+\lg (1 / x)=0 \) för varje positivt värde på \( x \). Har hon rätt? Nej Ja

Ask by Harrington Frank. in Sweden
Jan 21,2025

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Answer

Gunilla har rätt. Uttrycket \( \lg x + \lg \left( \frac{1}{x} \right) = 0 \) gäller för alla positiva värden på \( x \).

Solution

För att avgöra om Gunilla har rätt, låt oss analysera uttrycket \( \lg x + \lg \left( \frac{1}{x} \right) \). Vi kan använda logaritmlagarna för att förenkla uttrycket. Enligt logaritmlagarna gäller att: \[ \lg a + \lg b = \lg (a \cdot b) \] I vårt fall kan vi skriva om \( \lg \left( \frac{1}{x} \right) \) som: \[ \lg \left( \frac{1}{x} \right) = \lg (1) - \lg (x) \] Eftersom \( \lg (1) = 0 \), får vi: \[ \lg \left( \frac{1}{x} \right) = 0 - \lg (x) = -\lg (x) \] Nu kan vi sätta in detta i vårt ursprungliga uttryck: \[ \lg x + \lg \left( \frac{1}{x} \right) = \lg x - \lg x = 0 \] Detta visar att \( \lg x + \lg \left( \frac{1}{x} \right) = 0 \) är sant för varje positivt värde på \( x \). Så Gunilla har rätt. Svaret är **Ja**.

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Beyond the Answer

Ja, Gunilla har rätt! Uttrycket \( \lg x + \lg (1/x) \) kan förenklas till \( \lg x - \lg x = 0 \) enligt logaritmlagarna. Eftersom logaritmen av en kvot blir skillnaden av logaritmerna, innebär detta att summan alltid kommer att vara noll så länge \( x \) är positivt. Det är också värt att nämna att logaritmer är ett kraftfullt verktyg inom matematik och används ofta inom olika områden, från att lösa exponentiella ekvationer till att analysera tillväxttakter i vetenskapen och teknologin. Logaritmer hjälper oss att hantera mycket stora eller små tal på ett mer hanterbart sätt!

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