What is the solution to log5?

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You solve logarithms with exponentials, here is the relationship, assuming you mean the natural logarith of 5.
:
The base of a natural logarithm is e (the number e, 2.718...)
therefore:
log_b(x) = y
b^ŷ = x
log_e 5 = y
e^ŷ = 5
or if you want to use the numerical equivalent of e to three decimals:
2.718^ŷ = 5

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There is no way other than approximation by series or
iterative methods, because it is an irrational number.

0.6989700043360188047862611052755069732318101185378914586895725388728918107255754905130727478818138279593155228085690046209052321188664764940003076662953044249354970357458065973381802656883970564988160971018214173828455604681380...

The "solution" above uses e, but there is no way to find
e other than by approximation by series or iterative
methods, because it is an irrational number. It only uses
e to 3 decimal places.  
Edwin

You can put this solution on YOUR website!
.
Without using a calculator, you may look to trigonometric tables
(as your grandfather and grandmother did in pre-computer era)
or to the Internet.

Another way could be using#log2=0.3010#(again for this we need log tables) as#log5=log(10/2)=log10-log2=1-0.3010=0.6990#

In fact one need to remember log (to the base#10#) for first ten numbers and it makes things lot easier. Note that while#log1=0#,#log10=1#. In fact, you just need to know#log2=0.3010#,#log3=0.4771#and#log7=0.8451#and then all logs can be worked out using these as

#log4=2log2#,

#log5=1-log2#,

#log6=log2+log3#,

#log8=3log2#and

#log9=2log3#

If you want to go further, you need to remember#log11=1.0414#,#log13=1.1139#,#log17=1.2304#and#log19=1.2788#, rest can be easily calculated. For example#log14=log2+log7=1.1461#.

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