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3^x= 3*2^x solve this equation​

User RealPro
by
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1 Answer

7 votes

In the equation


3^x = 3\cdot 2^x

divide both sides by
2^x to get


(3^x)/(2^x) = 3 \cdot (2^x)/(2^x) \\\\ \implies \left(\frac32\right)^x = 3

Take the base-3/2 logarithm of both sides:


\log_(3/2)\left(\frac32\right)^x = \log_(3/2)(3) \\\\ \implies x \log_(3/2)\left(\frac 32\right) = \log_(3/2)(3) \\\\ \implies \boxed{x = \log_(3/2)(3)}

Alternatively, you can divide both sides by
3^x:


(3^x)/(3^x) = (3\cdot 2^x)/(3^x) \\\\ \implies 1 = 3 \cdot\left(\frac23\right)^x \\\\ \implies \left(\frac23\right)^x = \frac13

Then take the base-2/3 logarith of both sides to get


\log_(2/3)\left(2/3\right)^x = \log_(2/3)\left(\frac13\right) \\\\ \implies x \log_(2/3)\left(\frac23\right) = \log_(2/3)\left(\frac13\right) \\\\ \implies x = \log_(2/3)\left(\frac13\right) \\\\ \implies x = \log_(2/3)\left(3^(-1)\right) \\\\ \implies \boxed{x = -\log_(2/3)(3)}

(Both answers are equivalent)

User Skjerns
by
8.4k points

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