logarithm question
solve \(\large \color{black}{\begin{align} \log_{10}(x^{3}+5)=3\log_{10}(x+2)\hspace{.33em}\\~\\ \end{align}}\)
\(\large \color{black}{\begin{align} & (a.)\ \dfrac{-2+\sqrt2}{2} \hspace{.33em}\\~\\ & (b.)\ \dfrac{-2-\sqrt2}{2} \hspace{.33em}\\~\\ & (c.)\ \text{both a.) and b.) } \hspace{.33em}\\~\\ & (d.)\ \text{none of these} \hspace{.33em}\\~\\ \end{align}}\)
\[\log_{10}(x^3 + 5) = \log_{10}{(x+2)^3}\]\[\Rightarrow x^3 + 5 = x^3 + 6x^2 + 12x + 8\]\[\Rightarrow 6x^2 + 12x + 3 = 0 \]Choose the solution for which both logarithms are defined.
both a.) and b.)
?
Yeah, I guess so.
my book has given option d.)
If you know that your answer is correct, why care about the book?
lol i am not sure
i dont have brains
You're correct. Don't worry.
all options of book is correct except this ? suspictious
Book bhi ek insaan ne hi likhi hai. As I said, your answer is correct.
you may double check with wolfram
wolfram gave this http://www.wolframalpha.com/input/?i=solve+%5Clog_%7B10%7D%28x%5E%7B3%7D%2B5%29%3D3%5Clog_%7B10%7D%28x%2B2%29+over+reals&dataset=
So you're right.
ok thnx, by the way is there a condition like \(x^{3}+5>0\) and \((x+2)^{3}>0\)
Yes, which is what I meant that the logarithms should be defined. Both roots satisfy both conditions.
how do i find value of \(5^{1/3}\) by pen paper to check
You don't need to find the value of \(5^{1/3}\). Plug in \(1.4 \) for \(\sqrt 2\) and that should almost always work.
y i dont need that value
You need to check if \(x^3 + 5 > 0\) and \(x+2>0\). The second one is easier than the first. The solutions you get are\[\frac{-2 + \sqrt 2}{2} \approx - 0.3 , \frac{-2 - \sqrt 2}{2} \approx -1.7 \]The second one is simple. The first condition, well, is satisfied the first root. The second root is a closer call but it satisfies the condition.
satisfied for the first root*
You are the victim of a typo.... it happens quite a bit in math books because it is very hard to check the answers....
no typo for x^3+5>0
He's talking about the answer-key.
oh lol
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