How do you solve #ln(x+1)-lnx=5#?

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Answer 1 · 2017-06-09T14:01:05

See below.

When we subtract two natural logs (or two logs in general), we divide the expressions on the inside.

#ln(x+1)-lnx=5#

#ln((x+1)/x)=5#

#e^5=(x+1)/x#

#xe^5=x+1#

#xe^5-x=1#

#x(e^5-1)=1#

#x=1/(e^5-1)#

Answer 2 · 2017-06-09T14:02:37

I got: #x=1/(e^5-1)=0.006783#

We can use a property of logs:

#logx-logy=log(x/y)#

and write:

#ln((x+1)/x)=5#

use the definition of (natural) log:

#(x+1)/x=e^5#

rearrange:

#x+1=xe^5#

#x(e^5-1)=1#

#x=1/(e^5-1)=0.006783#