Either use factoring or the quadratic formula to solve the given equation.
(log10 x)^2 + log10 x = 6

Question

Either use factoring or the quadratic formula to solve the given equation.
(log10 x)^2 + log10 x = 6

jdoe0001 · Answer

$\bf (log_{10}x)^2+log_{10}x=6 ?$

solomonzelman · Answer

$\large\color{black}{ \displaystyle {\rm let~}~~~~a= (\log_{10}~x) }$ just to visualize it.

solomonzelman · Answer

$\large\color{black}{ \displaystyle  a^2+a=6  }$

solomonzelman · Answer

then subtract 6 from both sides, and factor

anonymous · Answer

so you would get 3 and -2

solomonzelman · Answer

a^2 + a = 6
a^2 + a - 6 = 0
(a + 3)(a - 2) = 0

a=-3  or  2

solomonzelman · Answer

(( you were almost right ))

solomonzelman · Answer

So,

$\large\color{black}{ \displaystyle \log_{10}x=-3  }$

and

$\large\color{black}{ \displaystyle \log_{10}x=2  }$

solomonzelman · Answer

which one does not work ?

anonymous · Answer

-3

solomonzelman · Answer

yup

solomonzelman · Answer

So, all you are left with is:    $\large\color{black}{ \displaystyle \log_{10}x=2  }$

anonymous · Answer

so the answers would be log10x=2...can you simplify that at all?

solomonzelman · Answer

yes, you can solve for x.

solomonzelman · Answer

$\large\color{black}{ \displaystyle \log_{A}B=C~~~~~\Longrightarrow~~~~~~A^C=B  }$

anonymous · Answer

so 10^3

solomonzelman · Answer

$\large\color{black}{ \displaystyle \log_{10}x=2~~~~~\Longrightarrow~~~~~~~~x=? }$

solomonzelman · Answer

retry

solomonzelman · Answer

(or typo)

anonymous · Answer

10^2

solomonzelman · Answer

yup

anonymous · Answer

so 1000 is my answer

solomonzelman · Answer

10^2=1000 ?

solomonzelman · Answer

are you sure ?

anonymous · Answer

100 sorry typo

solomonzelman · Answer

yes, I can show how to make the keyboard slower (less "sensitive")....

solomonzelman · Answer

anyway.... x=100
(that is your only solution that works )

anonymous · Answer

my homework says it doesnt work...hmmmm

solomonzelman · Answer

$\large\color{black}{ \displaystyle (\log_{10}x)^2+(\log_{10}x)=6  }$

$\large\color{black}{ \displaystyle (\log_{10}100)^2+(\log_{10}100)=6  }$

$\large\color{black}{ \displaystyle (\log_{10}10^2)^2+(\log_{10}10^2)=6  }$

$\large\color{black}{ \displaystyle (2\log_{10}10)^2+(2\log_{10}10)=6  }$

$\large\color{black}{ \displaystyle (2)^2+(2)=6  }$

$\large\color{black}{ \displaystyle 4+2=6  }$

solomonzelman · Answer

I don't see any fault in x being 100. and it is 100....

solomonzelman · Answer

tell your teachers that THEY don't work !!!

solomonzelman · Answer

wait, I think there is another answer...

solomonzelman · Answer

$\large\color{black}{ \displaystyle (\log_{10}x)=-3  }$

$\large\color{black}{ \displaystyle x=10^{-3}=1/1000  }$

solomonzelman · Answer

yes, that one can also work....

((you can not take a log of a negative number, but it can be equal to a negative number))

solomonzelman · Answer

So, 2 solutions
x=1/100
x=1/1000