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Mathematics 16 Online
OpenStudy (anonymous):

6^x-40 * 6^-x=6

OpenStudy (mertsj):

\[6^{x-40}*6^{-5}=6\]

OpenStudy (mertsj):

is that it?

OpenStudy (anonymous):

So sorry, no, I will clarify \[6^{x}-40 * 6^{-x}=6\]

OpenStudy (anonymous):

you factor out 6^x

OpenStudy (mertsj):

Let y = 6^x

OpenStudy (mertsj):

\[y-\frac{40}{y}=6\]

OpenStudy (mertsj):

\[\frac{y^2-40}{y}=6\]

OpenStudy (mertsj):

\[y^2-40=6y\]

OpenStudy (mertsj):

\[y^2-6y-40=0\]

OpenStudy (mertsj):

Factor, solve, resubstitute.

OpenStudy (anonymous):

(y-10)(y+4)= 0?

OpenStudy (mertsj):

yes

OpenStudy (anonymous):

ok so now that it is factored, how would I solve it?

OpenStudy (anonymous):

y= 10 and y =-4 two solutions

OpenStudy (anonymous):

thats what I would plug in for my exponents? my problem is asking "x = ?"

OpenStudy (anonymous):

so y= 6^x

OpenStudy (anonymous):

6^x = 10 and 6^x= -4, since 6^x cannot = -4 then solve for 6^x=10

OpenStudy (anonymous):

should be around 1.3

OpenStudy (anonymous):

to solve for x you take the log of both sides ... log 6^x = log 10 then put the x in front of the log 6 according to log rules. you get x log 6 = log 10. divide both sides by log 6 and you get log10/log 6 = 1.285 = x

OpenStudy (anonymous):

byeee

OpenStudy (anonymous):

Ok, I think I get it, thank you

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