In the given equation the value of x is 3 2-5 9 - 3x-272 x-5A- i- 1-54 - 109- ii- 1-6 - 10-7B- i- 1-54 - 103- ii- 1-6 - 107

The correct option is B i. 1.54 x 103, ii. 1.6x107 Given 9 × 3^x = (27)^2x 5 ⇒ 3^2 × 3^x = (3^3)^2x 5 ⇒ 3^x + 2 = 3^3(2x 5) Equating the powers x + 2 = ...

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Byju's Answer

Standard IX

Mathematics

Comparison of Quantities Using Exponents

In the given ...

Question

In the given equation the value of $x$ is $3 \frac{2}{5}$

$9 \times 3^{x} = (27)^{2 x - 5}$

i. 1.54 x 10³, ii. 1.6x10⁷

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i. 1.54x10⁹, ii. 1.6x10^-7

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Solution

The correct option is A
i. 1.54 x 10³, ii. 1.6x10⁷

Given

$9 \times 3^{x} = (27)^{2 x - 5}$

$\Rightarrow$ $3^{2} \times 3^{x} = (3^{3})^{2 x - 5}$

$\Rightarrow$ $3^{x + 2} = 3^{3 (2 x - 5)}$

Equating the powers

$x + 2 = 6 x - 15$

$\Rightarrow x = 3 \frac{2}{5}$

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Similar questions

Q. Solve:

9 \times 10^{9} \times \frac{1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{(5.3 \times 10^{- 11})^{2}} .

Q. Find the values of

x .

(i)

3^{x} \times 3^{- 2} = 3^{2}

(ii)

5^{- 3} \div 5^{x} = 5^{2}

(iii)

{(\frac{11}{5})}^{- 5} \times {(\frac{11}{5})}^{- 2} = {(\frac{11}{5})}^{x - 9}

Q. Find the number from each of the expanded form:
(i) 7 × 10⁸ + 3 × 10⁵ + 7 × 10² + 6 × 10¹ + 9
(ii) 4 × 10⁷ + 6 × 10³ + 5

Question 109

In the given figures (i) and (ii), find the values of a, b and c.

The standard exponential form of 1,540,000,000 is 1.54 x 10 $^{9}$ .

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In the given equation the value of x is 325 9×3x = (27)2x−5

The correct option is A i. 1.54 x 103, ii. 1.6x107 Given 9×3x = (27)2x−5 ⇒ 32×3x=(33)2x−5 ⇒ 3x+2=33(2x−5) Equating the powers x+2=6x−15 ⇒ x=325

In the given equation the value of $x$ is $3 \frac{2}{5}$

$9 \times 3^{x} = (27)^{2 x - 5}$

The correct option is A
i. 1.54 x 10³, ii. 1.6x10⁷

Given

$9 \times 3^{x} = (27)^{2 x - 5}$

$\Rightarrow$ $3^{2} \times 3^{x} = (3^{3})^{2 x - 5}$

$\Rightarrow$ $3^{x + 2} = 3^{3 (2 x - 5)}$

Equating the powers

$x + 2 = 6 x - 15$

$\Rightarrow x = 3 \frac{2}{5}$