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Expert: Sherman D. - 2/27/2008


Question
How do I solve for:  log(x+5)+log(3x-7)=2

Answer
log(x + 5) + log(3x - 7) = 2
log((x + 5)(3x - 7)) = 2
log(3x^2 - 7x + 15x - 35) = 2
log(3x^2 + 8x - 35) = 2
3x^2 + 8x - 35 = 10^2
3x^2 + 8x - 35 = 100
3x^2 + 8x - 135 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-8 ± sqrt(8^2 - 4(3)(-135)))/(2(3))
x = (-8 ± sqrt(64 + 1620))/6
x = (-8 ± sqrt(1684))/6
x = (-8 ± sqrt(4 * 421))/6
x = (-8 ± 2sqrt(421))/6
x = (-4/3) ± (1/3)sqrt(421)

since you can't have a negative log.

x = (-4/3) + (1/3)sqrt(421)

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Sherman D.

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I can answer questions dealing in mathematics of all kinds except for Physics and Calculus, but i can answer questions in Pre-Calculus and Chemistry. I can also answer questions in Recipes of all kinds. I can find games cheats/walkthroughs, but i can`t find a specific game online or offline. I can also do history and recipes for alcoholic beverages.

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Mathematics, Recipes, History, and Games.

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High School graduated. I graduated with honors, and i was in Beta Club for a year and a half.

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Principle's list and A and B honor roll in high school only.

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