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Home Q & A Board Science-Related Homework Help Mathematics
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fodog414 fodog414
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9 years ago
So I think I found the inverse but I'm not sure on how to put it in the form of log_b(A)+C could someone explain it step by step if they don't mind I'd love to learn on how to actually do it and apply it Slight Smile

Not really love or really want to apply it but just be able to do well on the final Slight Smile Slight Smile

If g(x)=1/4 (3)^{x+ 5}, calculate  g^-1(x) = log_b(A)+C, where b =  , A =  , C =

Thank you for taking the minute !!
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padrepadre
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9 years ago
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fodog414 Author
wrote...
#2
9 years ago Edited: 9 years ago, fodog414
Thanks padre
Post Merge: 9 years ago

I just don't understand on how to put the inverse in log_b(A)+C Frowning Face
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#3
Staff Member
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9 years ago
Me neither Confounded Face Any updates?
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fodog414 Author
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#4
9 years ago
Yes =)

So I found the inverse of 1/4(3)^x+5
Which =(ln(4x)-5ln(3))/ln(3)

simplified (ln(4x)-5ln(3))/ln(3)
to get (ln(4x))/(ln(3))-5


more simplification results with

 ln(4x)-ln(3)-5

I then got C for -5

A for 4x

b for 3
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#5
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9 years ago
Thanks for the update.
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