How do you find the inverse of #f(x)=2-3 log_4(x+1)#?

Question

Xavier Alonso · Accepted Answer

undefined To find the inverse of $ f(x) = 2 - 3 \log_4(x+1) $:

1. Replace $ f(x) $ with $ y $.
2. Swap $ x $ and $ y $: $ x = 2 - 3 \log_4(y+1) $.
3. Solve for $ y $.
4. First, isolate the logarithmic term: $ 3 \log_4(y+1) = 2 - x $.
5. Divide both sides by 3: $ \log_4(y+1) = \frac{2 - x}{3} $.
6. Rewrite in exponential form: $ y+1 = 4^{\frac{2 - x}{3}} $.
7. Subtract 1 from both sides: $ y = 4^{\frac{2 - x}{3}} - 1 $.

Therefore, the inverse of $ f(x) $ is $ f^{-1}(x) = 4^{\frac{2 - x}{3}} - 1 $.

James Allen · Answer

#y=4^((2-x)/3)-1#

Rewrite as #y=2-3log_4(x+1)#.

Swap the #x# and #y#.

#x=2-3log_4(y+1)#

Solve for #y#.

#x-2=-3log_4(y+1)#

#(2-x)/3=log_4(y+1)#

#4^((2-x)/3)=y+1#

#y=4^((2-x)/3)-1#

This can be solved another way. Return to #x-2=-3log_4(y+1)#.

#x-2=log_4(y+1)^-3#

#4^(x-2)=(y+1)^-3#

Raise both sides to the #-1/3# power.

#4^(-1/3(x-2))=y+1#

Again, we get:

#y=4^((2-x)/3)-1#

Elias Ackerman · Answer

#f^(-1)(x) = 4^((2-x)/3)-1#

To find the inverse of a function #f(x)# we can let #y = f(x)# and then solve for #x# to obtain #x = f^(-1)(y)# (To see why the inverse function is obtained, substitute #f(x)# back in for #y# and note that the new function applied to the original returns #x#).

Applying this here:

Let #y = f(x) = 2-3log_4(x+1)#

#=> 3log_4(x+1) = 2 - y#

#=> log_4(x+1) = (2-y)/3#

#=>4^(log_4(x+1)) = 4^((2-y)/3)#

#=> x+1 = 4^((2-y)/3)#

#=> x = 4^((2-y)/3)-1#

Thus we have #f^(-1)(y) = 4^((2-y)/3)-1#

meaning

#f^(-1)(x) = 4^((2-x)/3)-1#