To solve this equation, it's confusing
No.50
#Help me
sweet heath
swift inletBOT
dusk swallow
Do you mean $\left(\log_2{x}\right)^2 + \left(\log_2{2x}\right)^2 = 5$?
spiral ravenBOT
Jay
dusk swallow
If so, note that $\log_2{2x}$ can be simplified by log
laws to become $\log_2{2} + \log_2{x} = 1 + \log_2{x}$.
And then, if you substitute $U = \log_2{x}$, you should
get something that you will find easier to solve...
spiral ravenBOT
Jay
sweet heath
The first log2 x² the ² is on x alone
dusk swallow
Then the first term is $\log_2{x^2} = 2\log_2{x}$ by the log laws
spiral ravenBOT
Jay
dusk swallow
And so, the equation is
\begin{align*} \log_2{\left(x^2\right)} + \left(\log_2{2x}\right)^2 &= 5 \ 2\log_2{x} + \left(\log_2{2}+\log_2{x}\right)^2 &= 5 \qquad \text{using log laws} \ 2U + (1+U)^2 &= 5 \qquad \text{on substituting $U=\log_2{x}$} \end{align*}
spiral ravenBOT
Jay
atomic geode
$\left(\log_2{x}\right)^2$ doesnt it look like this?
spiral ravenBOT
ЯεтιяεĐ
dusk swallow
$\left(\log_2{x}\right)^2$ doesnt it look like this?
That's what I read it as originally, but @sweet heath is the OP and has said the squared was on the x only...
atomic geode
the brackets clearly tell the other
dusk swallow
Taken as $\left(\log_2{x}\right)^2 + \left(\log_2{2x}\right)^2 = 5$,
the equation becomes $U^2 + (1+U)^2 = 5$
spiral ravenBOT
Jay
dusk swallow
Either way, the question becomes a quadratic in $U$ after substitution
spiral ravenBOT
Jay
spiral ravenBOT
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dusk swallow
Btw how do you use that TEX app on discord
You can see the code I have been typing, except that some of the characters disappear... just to add to the challenge!
sweet heath
.close