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Simplifying exponents (Please Help!)
- Thread starter zee
- Start date
when you multiply 2 numbers with exponents:
\(\displaystyle a^b c^d = (ac)^{b + d}\)
when you take a number with an exponent an raise it to another exponent:
\(\displaystyle \left( {a^b } \right)^c = a^{bc}\)
See if you can use these rules to solve the question.
\(\displaystyle a^b c^d = (ac)^{b + d}\)
when you take a number with an exponent an raise it to another exponent:
\(\displaystyle \left( {a^b } \right)^c = a^{bc}\)
See if you can use these rules to solve the question.
jsbeckton said:when you multiply 2 numbers with exponents:
\(\displaystyle a^b c^d = (ac)^{b + d}\)
Unfortunately, this first "rule" is not valid. You can only add the exponents if the bases are alike. It appears that a does not equal c here.
The rule actually is a^b(a^c) = a^(b+c)
when you take a number with an exponent an raise it to another exponent:
\(\displaystyle \left( {a^b } \right)^c = a^{bc}\)
See if you can use these rules to solve the question.
Perhaps I should have put "5a" and "a" instead of a and c. However, the bases are the same. They are both in terms of "x".
\(\displaystyle \begin{array}{l}
y = \left[ {(5x^2 )(x^{ - 2} )} \right]^2 \\
y = \left[ {5x^0 } \right]^2 \\
y = \left[ 5 \right]^2 \\
y = 25 \\
\end{array}\)
You can add the exponents in this situation.
\(\displaystyle \begin{array}{l}
y = \left[ {(5x^2 )(x^{ - 2} )} \right]^2 \\
y = \left[ {5x^0 } \right]^2 \\
y = \left[ 5 \right]^2 \\
y = 25 \\
\end{array}\)
You can add the exponents in this situation.