How do you find the horizontal asymptote of 9(1/5)^x
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# How do you find the horizontal asymptote of 9(1/5)^x

[From: ] [author: ] [Date: 11-05-25] [Hit: ]
any kind of help would be very appreciated!since log(1/5) is a negative number, as x increases in postive direction,in fact,but the antilog oflog y = -inf isy = zero.so y -> 0,......
I am totally lost on this question...any kind of help would be very appreciated!

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y = 9 * (1/5)^x
take the logarithm

log y = log 9 + x*log (1/5)
since log(1/5) is a negative number, as x increases in postive direction, log y decreases
in fact, log y will go all the way to -inf as x goes to +inf (inf = infinity)

but the antilog of log y = -inf is y = zero.

so y -> 0, or zero is the horizontal asymptote
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