How do you solve #-9\cdot 17^ { - 4n - 8} - 4= - 40#?

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Answer 1 · 2017-12-04T01:08:16

#n=(log4/log17+8)/-4~=(.4893+8)/-4~=-2.1223#

#-9xx17^(-4n-8)-4=-40#

Let's move all the non-#n# terms to the right:

#17^(-4n-8)=(-40+4)/-9=4#

Now we can take the log of both sides:

#log(17^(-4n-8))=log4#

Remember that #logx^a=alogx#

#(-4n-8)log17=log4#

Divide by #log17#:

#-4n-8=log4/log17#

#n=(log4/log17+8)/-4~=(.4893+8)/-4~=-2.1223#