Solve for x: Logarithmic Equation with Multiple Terms - Step by Step Guide

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The discussion revolves around solving the logarithmic equation log x^2 + log x - log 2.7 = log 10. Participants clarify that when adding logarithms, the correct approach is to combine them into a single logarithm using multiplication, leading to log(x^3) = log(27). The confusion arises from misapplying logarithmic properties, particularly in manipulating the terms. Ultimately, the correct interpretation simplifies the equation to find x, emphasizing the need to correctly apply logarithmic rules. The conversation highlights the importance of understanding logarithmic identities for solving such equations effectively.
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Homework Statement



log x2 + log x - log 2.7 = log 10
solve for x

Homework Equations





The Attempt at a Solution


this one i really don't know how to do. I can't find any examples, but this is what i tried (youre going to be like, what? but i had to attempt it.) I think when logs are in addition format, youre supposed to multiply them? does that make sense? i really need help on this one. :

2 log x + log x = log 10 + log 2.7
2 log x2 = log 27
4 log x = log 27

thats as far as i can get. i know its wrong. even one step in the right direction would help me.
 
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No, 2*log(x)+log(x) IS NOT 2*log(x^2). 2*log(x)+log(x)=3*log(x)=log(27). Try it from there. Or do log(x^2)+log(x)=log(x^3)=log(27).
 
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^
i had it that way the first time. i think i changed it because i got

3 log x = log 27
then to solve for x i got something that looks like this:

x = log 27/3 log

3 log? is it supposed to be 3 log 1?

if so, the answer is 0. i put 0 on my exam, although i set it up incorrectly. my prof would give a point for the correct answer and i got no points, so i don't think that's the answer.
 
"3 log" makes no sense. Go back a step to 3 log x = log 27.

What's another way to write 3 log x?
 
3*log(x)=log(x^3), doesn't it? 3log(x) is not 3*log*x. That's just plain silly, as Mark44 points out.
 
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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