Inverse of log functions f(x) = 3 log (4x - 7) - 8, etc.

[axrw]

How would you find the inverse of functions like:

f(x) = 3 log (4x - 7) - 8
f(x) = -log x

I can only seem to manage:

f(x) = log x

x = 10^y

y = 10^x

but on ones like the above I think I'm getting them wrong, the graphs don't look right when I put them into my calculator.

 

[Deleted member 4993]

Re: Inverse of logarithmic functions.

How would you find the inverse of functions like:

f(x) = 3 log (4x - 7) - 8

y + 8 = log [(4x-7)^3]

Now continue
f(x) = -log x

y = - log(x)

y = log(1/x)

x = log(1/y)

1/y = 10^x

y = 1/(10^x)

y = 10^(-x)
I can only seem to manage:

f(x) = log x

x = 10^y

y = 10^x

but on ones like the above I think I'm getting them wrong, the graphs don't look right when I put them into my calculator.

 

[axrw]

Thanks for replying, Subhotosh Khan.

f(x) = 3 log (4x - 7) - 8

y + 8 = log [(4x-7)^3]

x + 8 = log[(4y-7)^3]

(4y-7)^3 = 10^(x + 8)

4y-7 = 10^[(x + 8)/3]

4y = 10^[(x + 8)/3] + 7

y = (10^[(x + 8)/3] + 7)/4


Is that correct?

On a side note, I think I'm going to need to learn to use that Tex thing. As I slowly progress through my book my questions look more and more unreadable in ascii.

 

[Deleted member 4993]

Thanks for replying, Subhotosh Khan.

f(x) = 3 log (4x - 7) - 8

y + 8 = log [(4x-7)^3]

x + 8 = log[(4y-7)^3]

(4y-7)^3 = 10^(x + 8)

4y-7 = 10^[(x + 8)/3]

4y = 10^[(x + 8)/3] + 7

y = (10^[(x + 8)/3] + 7)/4....Correct may be I would write it as [10^{(x+8)/3}+7]/4

Is that correct?

On a side note, I think I'm going to need to learn to use that Tex thing. As I slowly progress through my book my questions look more and more unreadable in ascii.