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How do you find the inverse of #y=ln(8x + 1)#?

Answer 1

Cora Alexander

To find the inverse of the function ( y = \ln(8x + 1) ), follow these steps:

Begin by swapping the roles of ( x ) and ( y ) in the equation to change ( y ) to ( x ) and vice versa:

[ x = \ln(8y + 1) ]

Rewrite the equation in exponential form to isolate ( y ):

[ e^x = 8y + 1 ]

Subtract 1 from both sides:

[ e^x - 1 = 8y ]

Divide both sides by 8:

[ y = \frac{e^x - 1}{8} ]

So, the inverse of the function ( y = \ln(8x + 1) ) is ( y = \frac{e^x - 1}{8} ).

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Answer 2

Sophie Adams

#x=(e^y-1)/8#

If #y=ln(8x+1)# then (by definition) #color(White)("XXX")e^y=8x+1#

From this it follows that #color(White)("XXX")e^y-1 = 8x# and #color(White)("XXX")x=(e^y-1)/8#

It is important to remember: #color(White)("XXX")log_b a = c <=> b^c = a#

(and that #ln x = log_e x#)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.