Find x- if- i 42x - 1-32 ii -2x-3-16 iii -3-5 x-1 - 125-27 iv -2-3 x-1 - 27-8

(iv) ( √(2/3) )^x 1 = 27/8 ( 2/3 )^x 1/3 = 3^3/2^3 2^x 1/3/3^x 1/3 = 3^3/2^3 2^x 1/3× 2^3 = 3^x 1/3× 3^3 2^x 1/3 +3 =3^x 1/3 +3 x 1/3 +3 = 0 x 1/3 ...

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Byju's Answer

Standard IX

Mathematics

Normal Form to Standard Form

Find x, if: i...

Question

Find $x$ , if:

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

(ii) $\sqrt{2^{x + 3}} = 16$

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

(iv) ${(\sqrt[3]{\frac{2}{3}})}^{x - 1} = \frac{27}{8}$

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Solution

$i right parenthesis 4 to the power of 2 x end exponent equals 1 over 32 rightwards double arrow open parentheses 2 squared close parentheses to the power of 2 x end exponent equals 1 over 2 to the power of 5 rightwards double arrow 2 to the power of 4 x end exponent equals 2 to the power of negative 5 end exponent s o 4 x equals negative 5 x equals negative 5 over 4 i i right parenthesis square root of 2 to the power of x plus 3 end exponent end root equals 16 s q a r i n g space b o t h space s i d e space w e space g e t 2 to the power of x plus 3 end exponent equals 16 squared 2 to the power of x plus 3 end exponent equals 2 to the power of 8 s o x plus 3 equals 8 x equals 5 i i i right parenthesis open parentheses square root of 3 over 5 end root close parentheses to the power of x plus 1 end exponent equals 125 over 27 rightwards double arrow open parentheses square root of 3 over 5 end root close parentheses to the power of x plus 1 end exponent equals open parentheses 5 over 3 close parentheses cubed rightwards double arrow 3 to the power of begin display style fraction numerator x plus 1 over denominator 2 end fraction end style end exponent over 5 to the power of begin display style fraction numerator x plus 1 over denominator 2 end fraction end style end exponent equals 5 cubed over 3 cubed rightwards double arrow 3 to the power of fraction numerator x plus 1 over denominator 2 end fraction end exponent cross times 3 cubed equals 5 to the power of fraction numerator x plus 1 over denominator 2 end fraction end exponent cross times 5 cubed rightwards double arrow 3 to the power of fraction numerator x plus 1 over denominator 2 end fraction plus 3 end exponent equals 5 to the power of fraction numerator x plus 1 over denominator 2 end fraction plus 3 end exponent t h e n fraction numerator x plus 1 over denominator 2 end fraction plus 3 equals 0 fraction numerator x plus 1 over denominator 2 end fraction equals negative 3 x plus 1 equals negative 6 x equals negative 7$

$(i v) {(\sqrt[3]{\frac{2}{3}})}^{x - 1} = \frac{27}{8} (\frac{2}{3})^{\frac{x - 1}{3}} = \frac{3^{3}}{2^{3}} \frac{2^{\frac{x - 1}{3}}}{3^{\frac{x - 1}{3}}} = \frac{3^{3}}{2^{3}} 2^{\frac{x - 1}{3}} \times 2^{3} = 3^{\frac{x - 1}{3}} \times 3^{3} 2^{\frac{x - 1}{3} + 3} = 3^{\frac{x - 1}{3} + 3} \frac{x - 1}{3} + 3 = 0 \frac{x - 1}{3} = - 3 x - 1 = - 9 x = - 8$

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Similar questions

Find the values of x in each of the following:

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

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

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

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

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

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

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

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

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

Q. Find x, if:

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

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

Q. Solve the equations:

\frac{2 x^{3} - 3 x^{2} + x + 1}{2 x^{3} - 3 x^{2} - x - 1} = \frac{3 x^{3} - x^{2} + 5 x - 13}{3 x^{3} - x^{2} - 5 x + 13}

Q. Solve:

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

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Find x, if: (i) 42x=132 (ii) √2x+3=16 (iii) (√35)x+1=12527 (iv) (3√23)x−1=278

(iv) (3√23)x−1=278(23)x−13=33232x−133x−13=33232x−13×23=3x−13×332x−13+3=3x−13+3x−13+3=0x−13=−3x−1=−9x=−8

Find $x$ , if:

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

(ii) $\sqrt{2^{x + 3}} = 16$

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

(iv) ${(\sqrt[3]{\frac{2}{3}})}^{x - 1} = \frac{27}{8}$