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Number Systems

Short Answer Type

171.

Simplify :

$open parentheses 3 to the power of 1 third end exponent close parentheses to the power of 4$

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172.

Simplify :

$7 to the power of begin display style 1 fifth end style end exponent over 7 to the power of begin display style 1 third end style end exponent$

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173.
Simplify :

$13 to the power of bevelled 1 fifth end exponent.17 to the power of bevelled 1 fifth end exponent$

Solution is not provided.

$A n s. space space space space space 221 to the power of 1 fifth end exponent$

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174. Simplify:

fraction numerator left parenthesis 25 right parenthesis to the power of begin display style 3 over 2 end style end exponent cross times left parenthesis 343 right parenthesis to the power of begin display style 1 fifth end style end exponent over denominator 16 to the power of begin display style 5 over 4 end style end exponent cross times 8 to the power of begin display style 4 over 3 end style end exponent cross times 7 to the power of begin display style 3 over 5 end style end exponent end fraction

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175. Find the value of x if 2⁴ × 2⁵ = (2³)^x

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176.

Simplify:

$open parentheses 64 over 125 close parentheses to the power of negative 2 over 3 end exponent$

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177. Find the value of:

fraction numerator 4 over denominator left parenthesis 216 right parenthesis to the power of negative begin display style 2 over 3 end style end exponent end fraction minus fraction numerator 1 over denominator left parenthesis 256 right parenthesis to the power of negative begin display style 3 over 4 end style end exponent end fraction

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178. Simplify:

open parentheses 15 to the power of begin display style 1 third end style end exponent over 9 to the power of begin display style 1 fourth end style end exponent close parentheses to the power of negative 6 end exponent

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179. Simplify:

open parentheses 15 to the power of begin display style 1 fourth end style end exponent over 3 to the power of begin display style 1 half end style end exponent close parentheses to the power of negative 2 end exponent

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180. Prove that:

open parentheses 64 over 125 close parentheses to the power of negative 2 over 3 end exponent plus 1 over open parentheses begin display style 256 over 625 end style close parentheses to the power of begin display style 1 fourth end style end exponent plus fraction numerator square root of 25 over denominator cube root of 64 end fraction

103 Views

Previous Year Papers

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Short Answer Type

173. Simplify :

173.
Simplify :

$13 to the power of bevelled 1 fifth end exponent.17 to the power of bevelled 1 fifth end exponent$