Solve for z: 2.3(1.2)^5z = 3(4.1)^z

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The equation 2.3(1.2)^(5z) = 3(4.1)^z is analyzed for solutions using logarithmic properties. Initial attempts to simplify the equation led to confusion about whether solutions exist, with some participants asserting there are no solutions. However, others emphasized the importance of applying the logarithmic expansion rule correctly, suggesting that a solution does exist. Ultimately, the discussion clarifies that there is indeed a solution to the equation when approached properly. The key takeaway is the correct application of logarithmic properties to solve for z.
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Homework Statement



2.3(1.2)^5z = 3(4.1)^z solve for z

Homework Equations



the properties of logarithm

The Attempt at a Solution


okay I first divided the 3 with 2.3 and got 1.304 then i brough the 5z and the z down which gives me 5z log(1.2)=1.304 z log(4.1) then i plugged log of 1.2 and 4.1 into the calculator and got .079 and .613 which makes the equation 5z(.079)=1.034z(.613) i then distributed and got .395z=.8z which doesn't give me an answer is the answer no solutions or did i do sumthing wrong?
 
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Hint: log( A B^z ) = log(A) + log(B^z) = log(A) + z log(B)
 
silentsaber said:
...which doesn't give me an answer. Is the answer that there are no solutions or did I do something wrong?

Yes, there are no solutions to this equation. Unless you used z because you want complex solutions, in which case the properties of logarithms aren't the same.
 
qntty said:
Yes, there are no solutions to this equation.

There is most definitely a solution to this equation. Take giant_bog's advice in using that log expansion rule.
 
meiso said:
There is most definitely a solution to this equation. Take giant_bog's advice in using that log expansion rule.

You are correct. My mistake.
 
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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