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6^x-40 * 6^-x=6
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\[6^{x-40}*6^{-5}=6\]
is that it?
So sorry, no, I will clarify \[6^{x}-40 * 6^{-x}=6\]
you factor out 6^x
Let y = 6^x
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\[y-\frac{40}{y}=6\]
\[\frac{y^2-40}{y}=6\]
\[y^2-40=6y\]
\[y^2-6y-40=0\]
Factor, solve, resubstitute.
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(y-10)(y+4)= 0?
yes
ok so now that it is factored, how would I solve it?
y= 10 and y =-4 two solutions
thats what I would plug in for my exponents? my problem is asking "x = ?"
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so y= 6^x
6^x = 10 and 6^x= -4, since 6^x cannot = -4 then solve for 6^x=10
should be around 1.3
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
byeee
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Ok, I think I get it, thank you
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