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$4*3^x - 9*2^x = 5* 3^\frac x2 * 2^\frac x2$

I did not understand this equality how to solve it for $x$?

user494605
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  • See https://math.stackexchange.com/questions/1356236/what-is-the-solution-to-the-equation-9x-6x-2-cdot-4x-0 OR https://math.stackexchange.com/questions/1804249/solve-equation-4x-3-cdot6x2-cdot9x-0 – lab bhattacharjee Oct 23 '17 at 05:43

3 Answers3

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$$4*3^x - 9*2^x = 5* 3^\frac x2 * 2^\frac x2$$

$$4 - 9*\frac {2^x}{3^x} = 5* \frac {3^\frac x2}{3^x} * 2^\frac x2$$

$$4 - 9*\bigg(\frac 2 3\bigg)^x = 5* \frac{2^\frac x2}{3^\frac x2} $$

Substitute $a=(\frac 23 )^\frac x2$

$$4 - 9a^2 = 5a$$

Solve for a...

$$ 9a^2 +5a-4= 0$$

$$ (a+1)(9a-4)= 0$$

user173897
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user577215664
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Hint:

Solve $$4m^2-5mn-9n^2=0$$

by factorization.

Notice that $9-4=5$.

Siong Thye Goh
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$4*3^x - 9*2^x = 5* 3^\frac x2 * 2^\frac x2 $

The right side is $5\cdot 6^\frac x2 $.

The left side can be factored as

$4*3^x - 9*2^x =(2\cdot 3^\frac x2-3\cdot 2^\frac x2)(2\cdot 3^\frac x2+3\cdot 2^\frac x2) $.

By inspection, a solution is $x=4$. ($2\cdot 9= 18, 3\cdot 4 = 12, (18-12)(18+12) =6\cdot 30 =5\cdot 6^2$)

The left side is $3^x(4 - 9*(2/3)^x) \le 0 $ for $x \le 2$. Therefore $x > 2$.

If $x \ge 3$, $4 - 9*(2/3)^x \ge 4-8/3 =4/3 $ so, if $x \ge 3$, $5\cdot 6^\frac x2 \ge (4/3)3^x $ or $15/4 \ge (3/2)^{x/2} $ or $x \le 2\dfrac{\log(15/4)}{\log(3/2)} \lt 6.52 $.

Numerically, the only solution is $x=4$, and the left side appears to have a larger slope than the right side, so this would be the only solution.

I will leave it at this.

marty cohen
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