Mathematics
OpenStudy (anonymous):

Logarithms: Solve 6(10^2x) + 10^x - 2 =0

OpenStudy (anonymous):

Consider that 10^2x is really like saying (10^x) squared, or:$(10^x)^2$Now, rewrite the original problem by using a substitution of $y=10^x$Using that substitution, your original equation can be rewritten as$6y^2+y-2=0$which can be solved in a variety of manners (quadratic, factoring, etc). Just keep in mind when you do solve that the answer you're getting is y, not the original problem. Once you have y, replace y with the original 10^x and complete the problem.

OpenStudy (anonymous):

okay thank you :)

OpenStudy (anonymous):

yeah and unfortunately the solution is very ugly because this one does not factor over integers

myininaya (myininaya):

6y^2+4y-3y-2=0

OpenStudy (anonymous):

postive solution is $\frac{-5+\sqrt{37}}{6}$so answer is $\log(\frac{-5+\sqrt{37}}{6})$

myininaya (myininaya):

2y(3y+2)-(3y+2)=0 (2y-1)(3y+2)=0

OpenStudy (anonymous):

oh heavens you are right. i sovled $+y^2+10y-2=0$ like a dope

myininaya (myininaya):

y=1/2 y=-2/3 but u need solve for x remember y=10^x

OpenStudy (anonymous):

stop showing off your "splitting the middle term"!

myininaya (myininaya):

10^x=1/2 10^x=-2/3 satellite will tell u the rest

OpenStudy (anonymous):

satellite is beet red

OpenStudy (anonymous):

discard negative answer and write $\log(\frac{1}{2})$

myininaya (myininaya):

and dont forget you can't have log of a negative

myininaya (myininaya):

myininaya (myininaya):

lol

OpenStudy (anonymous):

i must be tired. jealous because polpak gets to be green and i don'

myininaya (myininaya):

satellite amistre is also green did you know?

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