Solve for x: 2^3+2^3+2^3+2^3=2^x

AI Thread Summary
To solve the equation 2^3 + 2^3 + 2^3 + 2^3 = 2^x, it's important to recognize that the left side can be simplified as a coefficient times 2^3. Instead of using logarithms, which can complicate the problem, it's more efficient to rewrite the sum as 4 * 2^3. This leads to the equation 4 * 2^3 = 2^x, where 4 can be expressed as 2^2. By equating the exponents, x can be determined as 5, since 2^5 equals 32. Using properties of exponents is the most straightforward approach to find x in this case.
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Homework Statement


Solve the following equation: 2^3+2^3+2^3+2^3=2^x[/B]

Homework Equations


log(a)^x=x*log(a)[/B]

The Attempt at a Solution


What i attempted was to log both sides, bring down the exponents, and summarize them. This left me with 12*log(2)=x*log(2). I then divide both sides by log(2) and get x=12, which is wrong. Please note that all of this was with logarithms with base number 10.
 
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Hi,

Where did you get your relevant equation ? It's dead (*) wrong, but visually close to the right one: ##\log (a^x) = x \log a##.

Taking logarithms of a sum is generally a bad idea. For a product you are better off. So make a product of the lefthand side.
And look at that. Maybe you don't need to take logarithms, but you might be able to use a simpler equation about exponentiation.

PS the term "summarizing" isn't all that mathematically sound... "sum" is what you mean. But you can't do that.

(*) [edit] well, dangerous is a better expression. It raises confusion between ##(\log a)^x## and ##(\log (a^x))##
 
I can see no reason to use logarithms. You are just trying to find x such that 2^x= 32. That should be elementary.
 
I second what BvU says. If this is an exercise of the properties of exponents, you should notice that 2^3 is a repeated sum...how many times is it repeated?
Rewrite it as a coefficient times 2^3, then write your coefficient as a power of 2.
Now, you will be able to use properties of exponents to quickly solve for x.

Otherwise, just add 8+8+8+8 and do what HallsofIvy suggested.

And if you are really feeling like using logarithms ... use log base 2.
 
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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