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Exponents with Like Bases

Question

${(\frac{2}{3})}^{2} \times {(\frac{2}{3})}^{5} = {(\frac{2}{3})}^{10}$
True
False

A

True

B

False

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Solution

Verified by Toppr

False.
$L H S = {(\frac{2}{3})}^{2} \times {(\frac{2}{3})}^{5}$
Using law of exponents, $a^{m} \times a^{n} = a^{m + n}$
Therefore, ${(\frac{2}{3})}^{2} \times {(\frac{2}{3})}^{5} = {(\frac{2}{3})}^{2 + 5} = {(\frac{2}{3})}^{7}$
$L H S \neq R H S$

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Similar Questions

Q1

{(\frac{2}{3})}^{2} \times {(\frac{2}{3})}^{5} = {(\frac{2}{3})}^{10}

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Q2

Question 59
${(\frac{7}{3})}^{2} \times {(\frac{7}{3})}^{5} = {(\frac{7}{3})}^{10}$

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Q3

If

2^{10} + 2^{9} \cdot 3^{1} + 2^{8} \cdot 3^{2} + \dots + 2 \cdot 3^{9} + 3^{10} = S - 2^{11},

then

S

is equal to:

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Q4

Find:

\frac{2}{3} + \frac{3}{3^{2}} + \frac{2}{3^{3}} + \frac{3}{3^{4}} + \frac{2}{3^{5}} + \frac{3}{3^{6}} + . . . .

to infinity.

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Q5

Simplify:
(i)

(- 4)^{5} \div (- 4)^{8}

(ii)

{(\frac{1}{2^{3}})}^{2}

(iii)

(- 3)^{4} \times {(\frac{5}{3})}^{4}

(iv)

{(\frac{2}{3})}^{5} \times {(\frac{3}{4})}^{2} \times {(\frac{1}{5})}^{2}

(v)

(3^{- 7} \div 3^{10}) \times 3^{- 5}

(vi)

\frac{2^{6} \times 3^{2} \times 2^{3} \times 3^{7}}{2^{8} \times 3^{6}}

(vii)

y^{a - b} \times y^{b - c} \times y^{c - a}

View Solution