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Suppose you have the product of two expressions: $ 2^5$ * $ 2^2$, the result will be : $ 2^{5+2}$ = $ 2^7$. This is because we know the exponent rule that if they have the same base, we can add the power. Is there a way to express the product of two expressions of different bases and powers into 1 expression with one base and power, like: $ 3^7$ * $ 4^8$ = $ a^b$

Monolica
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  • The fundamental theorem of arithmetic tells you no. – Andrew Chin Oct 28 '19 at 17:52
  • $3^7\cdot 4^8 = a^b$ where $a=3^7\cdot 4^8$ and $b=1$ – JMoravitz Oct 28 '19 at 17:52
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    The short answer is, "no, there isn't necessarily a convenient way to do this that is helpful and simplifies calculations." The longer answer is, of course there are ways you can write the result, but this can require you to leave the realm of the integers or can be particularly uninteresting like my previous example. – JMoravitz Oct 28 '19 at 17:54
  • @JMoravitz I can't see the condition on the integers. – user Oct 28 '19 at 18:03

2 Answers2

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You can use the fact that $x=a^{\log_a x}$, then $$ 3^7\times 4^8=(4^{\log_4 3})^7\times 4^8=4^{7\log_4 3}\times 4^8=4^{8+7\log_4 3}. $$

However, if you are asking whether there are non-trivial integer numbers $a$ and $b$, that $3^7\times 4^8=a^b$, then the answer is no.

Vasily Mitch
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  • Did you forget that $3^7\cdot 4^8$ is an integer and that you can raise an integer to the power $1$ and have it remain the same? There certainly do exist integers $a,b$ such that $3^7\times 4^8=a^b$. They may be trivial for this particular example, but not non-existant like your phrasing implies. – JMoravitz Oct 28 '19 at 17:56
  • Thanks. I overlooked this part. I have edited my answer – Vasily Mitch Oct 28 '19 at 17:58
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You can always do this by a simple change of base. Suppose you have the product $a^mb^n,$ and you want to write as a single expression involving just one base, then change the base $b$ to a power of $a,$ so that we have that $b=a^{\log_ab}.$ Thus, we find that $$a^mb^n=a^m(a^{\log_ab})^n=a^ma^{n\log_ab}=a^{m+n\log_ab}.$$

In your example you would now have $$3^74^8=3^72^{16}=3^73^{16\log_32}=3^{7+16\log_32}.$$

Allawonder
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