Determine the value of x in 233-23-6-232x-1-

Since bases are same we can equate the powers, [ ( 2 3 #160; ) ^ 3 ×( 2 3 #160; ) ^ 6 =( 2 3 #160; ) ^ 2x 1; ...

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Exponential Representation for Irrational Numbers

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Question

Determine the value of x in ${(\frac{2}{3})}^{3} \times {(\frac{2}{3})}^{- 6} = {(\frac{2}{3})}^{2 x - 1}$ .

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∣ ∣ ∣ \begin{matrix} \sqrt{2} & 2 \sqrt{3} 3 \sqrt{2} & \sqrt{3} \end{matrix} ∣ ∣ ∣

α, β, γ

x^{3} + 3 x^{2} + 2 = 0

\frac{α}{β + γ}, \frac{β}{γ + α}, \frac{γ}{α + β}

Find the values of x in each of the following:

$(i) 2^{5 x} \div 2^{x} = \sqrt[5]{2^{20}}$

$(i i) (2^{3})^{4} = (2^{2})^{x}$

$(i i i) {(\frac{3}{5})}^{x} {(\frac{5}{3})}^{2 x} = \frac{125}{27}$

$(i v) 5^{x - 2} \times 3^{2 x - 3} = 135$

$(v) 2^{x - 7} \times 5^{x - 4} = 1250$

$(v i) (\sqrt[3]{4})^{2 x + \frac{1}{2}} = \frac{1}{32}$

$(v i i) 5^{2 x + 3} = 1$

$(v i i i) (13)^{\sqrt{x}} = 4^{4} - 3^{4} - 6$

$(i x) {(\sqrt{\frac{3}{5}})}^{x + 1} = \frac{125}{27}$

(1 - y) (1 + 2 x + 4 x^{2} + 8 x^{3} + 16 x^{4} + 32 x^{5}) = (1 - y^{6})

\neq

x^{2} - 3 x + 2 = 0, x = 2, x = - 1

x^{2} + x + 1 = 0, x = 0, x = 1

x^{2} - 3 \sqrt{3} x + 6 = 0, x = \sqrt{3}, x = - 2 \sqrt{3}

x + \frac{1}{x} = \frac{13}{6}, x = \frac{5}{6}, x = \frac{4}{3}

2 x^{2} - x + 9 = x^{2} + 4 x + 3, x = 2, x = 3

x^{2} - \sqrt{2} x - 4 = 0, x = - \sqrt{2}, x = - 2 \sqrt{2}

a^{2} x^{2} - 3 a b x + 2 b^{2} = 0, x = \frac{a}{b}, x = \frac{b}{a}

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