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How can I show that (b^x-1)/x approaches Ln(b) as x approaches zero?

 Sep 21, 2014

Best Answer 

 #1
avatar+33603 
+5

Let z = (bx - 1)/x

 

Multiply both sides by x and add 1 to both sides of the result:

1 + zx = bx

 

Take (natural) logs of both sides

ln(1+zx) = xln(b)   using the property of logs that ln(mn) = n ln(m)

 

Divide both sides by x 

(1/x)*ln(1 + zx) = ln(b)

 

Use the same property of logs as we used above, but in reverse:

ln( (1 + zx)1/x ) = ln(b)

 

By definition, ez = limit as x goes to zero of  (1 + zx)1/x

So in the limit as x goes to zero we have

ln(ez) = ln(b)

or just z = ln(b)

 

or, using the definition of z

(bx - 1)/x = ln(b)

 Sep 22, 2014
 #1
avatar+33603 
+5
Best Answer

Let z = (bx - 1)/x

 

Multiply both sides by x and add 1 to both sides of the result:

1 + zx = bx

 

Take (natural) logs of both sides

ln(1+zx) = xln(b)   using the property of logs that ln(mn) = n ln(m)

 

Divide both sides by x 

(1/x)*ln(1 + zx) = ln(b)

 

Use the same property of logs as we used above, but in reverse:

ln( (1 + zx)1/x ) = ln(b)

 

By definition, ez = limit as x goes to zero of  (1 + zx)1/x

So in the limit as x goes to zero we have

ln(ez) = ln(b)

or just z = ln(b)

 

or, using the definition of z

(bx - 1)/x = ln(b)

Alan Sep 22, 2014

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