How do you solve 3^x+1=243 ?

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Answer 1 · 2017-12-28T22:13:56

See below.

Not sure whether this is #3^(x+1)=243# or #3^x+1=243#
#3^x+1=243# is what is written, but I'm suspecting it's supposed to be #3^(x+1)=243#

#3^(x+1)=243#

Taking logs of both sides:

#(x+1)ln(3)=ln(243)#

#x+1=ln(243)/ln(3)#

#x=ln(243)/ln(3)-1=ln(3^5)/ln(3)-1=(5ln(3))/ln(3)-1=4#

Answer 2 · 2018-08-05T02:42:45

#x=4#

We have the following:

#3^(x+1)=243#

We can rewrite #243# as #3^5#. We now have

#3^(x+1)=3^5#

Since we have the same base, we can equate the exponents. We get

#x+1=5#, which easily solves to

#x=4#

Hope this helps!