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

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.

okay thank you :)

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

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

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

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

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

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

stop showing off your "splitting the middle term"!

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

satellite is beet red

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

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

oh you already said it

lol

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

satellite amistre is also green did you know?

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