K/4 - 5k + 1 = 1 please

Mathematics

1 answer:

irinina [24]2 years ago

4 0

Answer:

-19k ≠ 0

No solutions

Step-by-step explanation:

k/4 - 5k + 1 = 1

4 (k/4 - 5k + 1 = 1)

k-20k + 4 = 4

-19k ≠ 0

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<img src="https://tex.z-dn.net/?f=%20%7B2%7D%5E%7Bx%7D%20%20-%20%20%20%7B3%7D%5E%7Bx%7D%20%20%3D%20%20%20%20%5Csqrt%7B%20%7B6%7D

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x = 0

Step-by-step explanation:

${2}^{x} - {3}^{x} = \sqrt{ {6}^{x} } - \sqrt{ {9}^{x } } \\ \\ {2}^{x} - {3}^{x} = \sqrt{ {6}^{x} } - \sqrt{ {( {3}^{2}) }^{x } } \\ \\ {2}^{x} - {3}^{x} = \sqrt{ {6}^{x} } - \sqrt{ {({3)}^{2x}}}\\ \\ {2}^{x} - \cancel{{3}^{x}} = \sqrt{ {6}^{x} } - \cancel{ {3}^{x} } \\ \\ {2}^{x} = \sqrt{ {6}^{x} } \\ squaring \: both \: sides \\ \\ {( {2}^{x} )}^{2} = {(\sqrt{ {6}^{x} })}^{2} \\ \\ {2}^{2x} = {6}^{x} \\ \\ {2}^{2x} = {(2 \times 3)}^{x} \\ \\ {2}^{2x} = {2 ^{x} \times 3}^{x} \\ \\ \frac{ {2}^{2x} }{ {2}^{x} } = 3^{x} \\ \\ {2}^{2x - x} = {3}^{x} \\ \\ {2}^{x} = {3}^{x} \\ \\ \frac{ {2}^{x} }{ {3}^{x} } = 1 \\ \\ \bigg( \frac{2}{3} \bigg)^{x} = 1 \\ \\ \implies \: x = 0 \\ \\ \because \: for \: x = 0 \\ \\ \bigg( \frac{2}{3} \bigg)^{0} = 1$