// Numbas version: exam_results_page_options {"name": "Indices: multiplying powers (algebraic)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Indices: multiplying powers (algebraic)", "tags": ["exponent", "exponents", "index", "index laws", "Index Laws", "indices", "multiplying", "nested power", "power", "power of a power", "powers"], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Simplify the following without the use of a calculator. Write your answer in index form using ^ to signify powers.

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2..6

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$\\left(d^\\var{powers1}\\right)^2$ = [[0]]

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Interpret the powers in expanded form and then write the result in index form.

\n
\n

Suppose you had to simplify $(5^3)^2$. Well cubed means there are three 5s all being multiplied, so we have

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\\[(5^3)^2=(5\\times 5\\times 5)^2\\]

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and squared means multiplied by itself (in this case the square is acting on the whole bracketed term) so we get

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\\[(5^3)^2=(5\\times 5\\times 5)^2=(5\\times 5\\times 5)\\times (5\\times 5\\times 5)\\]

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 but this is just six 5s all being multiplied together, that is

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\\[(5^3)^2=5^6.\\]

\n
\n

In general, we have $\\displaystyle(a^b)^c=a^{bc}$.

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Your answer is longer than necessary.

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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$\\displaystyle(j^\\var{powers2[0][0]})^\\var{powers2[1][0]}$ = [[0]]

\n

\n

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Interpret the powers in expanded form and then write the result in index form.

\n
\n

Suppose you had to simplify $(5^3)^2$. Well cubed means there are three 5s all being multiplied, so we have

\n

\\[(5^3)^2=(5\\times 5\\times 5)^2\\]

\n

and squared means multiplied by itself (in this case the square is acting on the whole bracketed term) so we get

\n

\\[(5^3)^2=(5\\times 5\\times 5)^2=(5\\times 5\\times 5)\\times (5\\times 5\\times 5)\\]

\n

 but this is just six 5s all being multiplied together, that is

\n

\\[(5^3)^2=5^6.\\]

\n
\n

In general, we have $\\displaystyle(a^b)^c=a^{bc}$.

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Your answer is longer than necessary.

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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Use the same approach you used in the above questions to simplify the following in index form.

\n


$\\displaystyle\\left(\\left(m^\\var{powers3[0][0]}\\right)^\\var{powers3[1][0]}\\right)^\\var{powers3[2][0]}$ = [[0]]

\n

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Interpret the powers in expanded form and then write the result in index form.

\n
\n

Suppose you had to simplify $((5^3)^2)^4$. We could start by applying the square

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 \\[((5^3)^2)^4=(5^3\\times 5^3)^4=(5^6)^4\\] 

\n

and then apply the fourth power

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\\[(5^6)^4=5^6\\times 5^6\\times 5^6\\times 5^6 = 5^{24}.\\]

\n
\n

Notice the above is the long way of just applying the rule  $\\displaystyle(a^b)^c=a^{bc}$ twice:

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\\[((5^3)^2)^4=(5^6)^4=5^{24}.\\]

\n

In fact, we could just multiply all the powers together in one go:

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\\[((5^3)^2)^4=5^{3\\times 2\\times 4}=5^{24}.\\]

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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$\\displaystyle\\left(\\left(n^\\var{nint}\\right)^\\var{ndec}\\right)^{\\var{num}/\\var{den}}$ = [[0]]

\n

\n

Note: If you want to use a fraction as a power you should use brackets to surround your power, for example, type 12^(2/3) for $12^\\frac{2}{3}$.

\n

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Multiply all the powers together as you have done in previous questions.

\n
\n

Note the answer can be written using a fractional power or a decimal power but a fractional power is more commonly used.

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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