How do you solve #2^x = 3#?

The inverse of an exponential function is the logarithm. In our question, we have a variable in the exponent of a base which is equal to #2#. The base of the exponent can be used to determine the base of the logarithm that we can use to solve the equation, in this case using that:

#log_2(2^x) = x#

We need to do the same thing to both sides of the equation, so it becomes:

#x=log_2(3)#

But now we need the #log_2# of #3# which some calculators can do, but many cannot. Instead we can use a logarithm of a base we know - almost all calculators can do both #log_10# and #log_e#. The #log_e# is also called the natural logarithm and is denoted either as #ln# or just log without a base, #log#.

To make this work, we need to know how to use a logarithm of a different base than that of the exponent. For this we use the change of base formula

#log_a(x) = log_n(x)/log_n(a)#

so our equation becomes

#x=log_2(3) = ln(3)/ln(2)~=1.099/0.6931 ~= 1.585#

Alternate approach to change of base

Another way to look at the change of base is to try using the natural logarithm on the left hand side of the equation, i.e.

#ln(2^x) = ln(3)#

We now have to ask ourselves, how do I express #2^x# as a power of my new base #x#? What if we change an #e# into a #2# using an exponent:

#e^c=2#

taking the natural log of this equation we also get

#c=ln(2)#

Then we could re-write the power of #2# as

#2^x = (e^c)^x=e^(cx)#

putting this into the equation above we get

#ln(e^(cx)) = ln(3)#

which gives

#cx=ln(3)#

We can substitute for #c# using an expression from above:

#ln(2)*x=ln(3)

and finally solving for #x# we get

#x=ln(3)/ln(2)#