Solve the following equations- i3x-1-27-34-ii42x-16-6-y-82-iii3x-1- 52y-3-225-iv8x-1-16y-2-and- 1-2 3-x- 1-4 3y-v4x-1- 0-53-2x- 1-8 x-vi-a-b- b-a 1-2x-where-a-b-are-distinct-positive-primes-

(i)3^x+1=27×3^4 3^x+1=3^3+4=3^7 ∴ x+1=7⇒ x=7 1=6 x=6 (ii)4^2x=(√(16))^ 6/y=(√(8))^2 4^2x=(√(8))^2=8^1/2× 2=8=(2)^3 (2^2)^2x=2^3⇒ 2^2× 2x=2^3 ⇒ 2^4x=2^3 Comparin ...

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Byju's Answer

Standard IX

Mathematics

Laws of Exponents for Real Numbers

Solve the fol...

Question

Solve the following equations:

$(i) 3^{x + 1} = 27 \times 3^{4} (i i) 4^{2 x} = (\sqrt[3]{16})^{\frac{- 6}{y}} = (\sqrt{8})^{2} (i i i) 3^{x - 1} \times 5^{2 y - 3} = 225 (i v) 8^{x + 1} = 16^{y + 2} a n d {(\frac{1}{2})}^{3 + x} = {(\frac{1}{4})}^{3 y} (v) 4^{x - 1} \times (0.5)^{3 - 2 x} = {(\frac{1}{8})}^{x} (v i) \sqrt{\frac{a}{b}} = {(\frac{b}{a})}^{1 - 2 x}, w h e r e a, b a r e d i s t i n c t p o s i t i v e p r i m e s .$

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Solution

$(i) 3^{x + 1} = 27 \times 3^{4} 3^{x + 1} = 3^{3 + 4} = 3^{7} ∴ x + 1 = 7 \Rightarrow x = 7 - 1 = 6 x = 6$

$(i i) 4^{2 x} = (\sqrt[3]{16})^{\frac{- 6}{y}} = (\sqrt{8})^{2} 4^{2 x} = (\sqrt{8})^{2} = 8^{\frac{1}{2} \times 2} = 8 = (2)^{3} (2^{2})^{2 x} = 2^{3} \Rightarrow 2^{2 \times 2 x} = 2^{3} \Rightarrow 2^{4 x} = 2^{3} C o m p a r i n g, w e g e t ∴ 4 x = 3 \Rightarrow x = \frac{3}{4} a n d (\sqrt[3]{16})^{\frac{- 6}{y}} = (\sqrt{8})^{2} = 2^{3} (\sqrt[3]{2^{4}})^{\frac{- 6}{y}} = 2^{3} {(2^{\frac{4}{3}})}^{\frac{- 6}{y}} = 2^{3} \Rightarrow 2^{\frac{4}{3} (\frac{- 6}{y})} = 2^{3} \Rightarrow 2^{\frac{- 8}{y}} = 2^{3} C o m p a r i n g, w e g e t ∴ \frac{- 8}{y} = 3 \Rightarrow y = \frac{- 8}{3}$

$(i i i) 3^{x - 1} \times 5^{2 y - 3} = 225 \Rightarrow 3^{x - 1} \times 5^{2 y - 3} = (15)^{2} = (3 \times 5)^{2} \Rightarrow 3^{x - 1} \times 5^{2 y - 3} = 3^{2} \times 5^{2} C o m p a r i n g, 3^{x - 1} = 3^{2} \Rightarrow x - 1 = 2 \Rightarrow x = 2 + 1 = 3 a n d, 5^{2 y - 3} = 5^{2} \Rightarrow 2 y - 3 = 2 \Rightarrow 2 y = 2 + 3 = 5 \Rightarrow y = \frac{5}{2} ∴ x = 3, y = \frac{5}{2}$

$(i v) 8^{x + 1} = 16^{y + 2} a n d {(\frac{1}{2})}^{3 + x} = {(\frac{1}{4})}^{3 y} (2^{3})^{x + 1} = (2^{4})^{y + 2} \Rightarrow 2^{3 x + 3} = 2^{4 y + 8} ∴ 3 x + 3 = 4 y + 8 \Rightarrow 3 x - 4 y = 8 - 3 = 5 . . . . (i) a n d {(\frac{1}{2})}^{3 + x} = {(\frac{1}{4})}^{3 y} {(\frac{1}{2})}^{3 + x} = {[{(\frac{1}{2})}^{2}]}^{3 y} = {(\frac{1}{2})}^{6 y} ∴ 3 + x = 6 y \Rightarrow x = 6 y - 3 . . . (i i) F r o m (i), 3 (6 y - 3) - 4 y = 5 \Rightarrow 18 y - 9 - 4 y = 5 14 y = 5 + 9 = 14 \Rightarrow y = \frac{14}{14} = 1 a n d x = 6 \times 1 - 3 = 6 - 3 = 3 ∴ x = 3, y = 1$

$(v) 4^{x - 1} \times (0.5)^{3 - 2 x} = {(\frac{1}{8})}^{x} \Rightarrow (2^{2})^{x - 1} \times {(\frac{1}{2})}^{3 - 2 x} = {(\frac{1}{2^{3}})}^{x} \Rightarrow 2^{2 x - 2} \times 2^{2 x - 3} = 2^{- 3 x} \Rightarrow 2^{4 x - 5} = 2^{- 3 x} \Rightarrow 4 x + 3 x = 5 \Rightarrow 7 x = 5 \Rightarrow x = \frac{5}{7}$

$(v i) \sqrt{\frac{a}{b}} = {(\frac{b}{a})}^{1 - 2 x} \Rightarrow {(\frac{a}{b})}^{\frac{1}{2}} = {(\frac{a}{b})}^{- 1 + 2 x} \Rightarrow \frac{1}{2} = - 1 + 2 x \Rightarrow 2 x = 1 + \frac{1}{2} = \frac{3}{2} \Rightarrow x = \frac{3}{2 \times 2} = \frac{3}{4} ∴ x = \frac{3}{4}$

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Similar questions

Q. Solve the following equations:
(i)

3^{x + 1} = 27 \times 3^{4}

(ii)

4^{2 x} = {(\sqrt[3]{16})}^{- \frac{6}{y}} = {(\sqrt{8})}^{2}

(iii)

3^{x - 1} \times 5^{2 y - 3} = 225

(iv)

8^{x + 1} = 16^{y + 2} and, {(\frac{1}{2})}^{3 + x} = {(\frac{1}{4})}^{3 y}

(v)

4^{x - 1} \times {(0.5)}^{3 - 2 x} = {(\frac{1}{8})}^{x}

(vi)

\sqrt{\frac{a}{b}} = {(\frac{b}{a})}^{1 - 2 x}

, where a and b are distinct primes

Solve the following equations for x:

(i) $7^{2 x + 3} = 1$

(ii) $2^{x + 1} = 4^{x - 3}$

(iii) $2^{5 x + 3} = 8^{x + 3}$

(iv) $4^{2 x} = \frac{1}{32}$

(v) $4^{x - 1} \times (0.5)^{3 - 2 x} = {(\frac{1}{8})}^{x}$

(vi) $2^{3 x - 7} = 256$

Q. Solve the following equations:
a)

5 (3 - x) - 2 (4 - 3 x) = 11 - 2 (x - 1)

6 - 4 (x + 3) = 2 (x - 1)

5 (1 - 2 x) + 2 (3 - x) = 3 (x + 4) + 14

Solve the equations:

(i) 5x = 3x + 24;

(ii) 8t + 5 = 2t − 31;

(iii) 7x − 10 = 4x + 11;

(iv) 4z + 3 = 6 + 2z;

(v) 2x − 1 = 14 − x;

(vi) 6x + 1 = 3(x − 1) + 7;

(vii) ;

(viii) ;

(ix) 3(x + 1) = 12 + 4 (x − 1);

(x) 2x − 5 = 3(x − 5);

(xi) 6(1 − 4x) + 7(2 + 5x) = 53;

(xii) 3(x + 6) + 2 (x + 3) = 64;

(xiii) ;

(xiv) .

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & x + 1 & x + 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = A x + B

where

A

and

B

are determinants of order

3

. Then

A + 2 B

is equal to

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Solve the following equations: (i)3x+1=27×34(ii)42x=(3√16)−6y=(√8)2(iii)3x−1×52y−3=225(iv)8x+1=16y+2 and (12)3+x=(14)3y(v)4x−1×(0.5)3−2x=(18)x(vi)√ab=(ba)1−2x,where a,b are distinct positive primes.