How do I evaluate this log expression?

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To evaluate the expression 5^(log_5(10) - 1), one can utilize the properties of exponents and logarithms. The expression can be rewritten using the identity 5^(a + b) = 5^a * 5^b, allowing for simplification. Alternatively, converting the constant 1 into a logarithmic form facilitates combining the logarithmic terms. Understanding that exponentials and logarithms are inverse functions is crucial in this process. Reviewing logarithmic properties can further aid in solving the expression effectively.
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Homework Statement


5^(log510-1)

Homework Equations


n/a

The Attempt at a Solution


I have no idea how to approach this.
 
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Well, you know what

5^{\log_5(x)}
is, don't you?

Then there are two ways to solve the question: you can split the sum in the power to a multiplication:
5^{a + b} = 5^a \cdot 5^b

or you can first write 1 as a logarithm (base 5), then combine the two logarithms into one.
 
Do you know that exponentials and logarithms are inverses of each other?
y = ax \Longrightarrow x = logay
Then, after substituting the x, y = a^{\log_a y}
So what is 5^{\log_5 x} ?

Perhaps you should review http://en.wikipedia.org/wiki/Logarithm" .
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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