3-4-5-x-4-6-x-1-dx-1-2-2x-1-find-the-value-of-x-Solution-4-5-x-3-4-6-x-2-2-x-k

Question Number 202636 by ibroclex_adex last updated on 30/Dec/23

\frac{^{3} \sqrt{4^{5 - x}}}{\int_{4}^{6} (x - 1) d x} = \frac{1}{2^{2 x - 1}}, find the value of x .

\underset{―}{Solution}

\frac{4^{\frac{5 - x}{3}}}{\int_{4}^{6} (\frac{x^{2}}{2} - x + k)} = \frac{1}{2^{2 x - 1}}

\frac{2^{2 ∙ \frac{5 - x}{3}}}{(\frac{6^{2}}{2} - 6 + k) - (\frac{4^{2}}{2} - 4 + k)} = \frac{1}{2^{2 x - 1}}

\frac{2^{\frac{10 - 2 x}{3}}}{\frac{36}{2} - 6 + k - \frac{16}{2} + 4 - k} = \frac{1}{2^{2 x - 1}}

\frac{2^{\frac{10 - 2 x}{3}}}{18 - 6 - 8 + 4} = \frac{1}{2^{2 x - 1}}

\frac{2^{\frac{10 - 2 x}{3}}}{8} = \frac{1}{2^{2 x - 1}} (Cross Multiply)

2^{2 x - 1} \times 2^{\frac{10 - 2 x}{3}} = 8 \times 1

2^{2 x - 1} \times 2^{\frac{10 - 2 x}{3}} = 2^{3}

2^{2 x - 1 + \frac{10 - 2 x}{3}} = 2^{3} (Since, the bases are equal . Then, we can equate the exponents)

2 x - 1 + \frac{10 - 2 x}{3} = 3 (Multiply each term by 3)

3 (2 x) - 3 (1) + 3 (\frac{10 - 2 x}{3}) = 3 (3)

6 x - 3 + 10 - 2 x = 9 (Collect Like Terms)

4 x + 7 = 9

4 x = 9 - 7

4 x = 2 (Divide Both Sides by 4)

\frac{4 x}{4} = \frac{2}{4}

∴ x = \frac{1}{2}