Question

3^x= 3*2^x solve this equation​

Answer 1

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)