Question 20616
Recall this about logarithms:
If {{{a^x = a^y}}}, the bases are the same, that is a = a, then the exponents are equal, therefore x = y.  Applying this to your problem:

{{{7^(logx) = 7^1}}} The bases are the same so the exponents are equal.

{{{log(x) = 1}}} Change to exponential form.
 Recall that log(x) is log to the base 10 of x:
Logarithmic form: y = log(x)...Exponential form: x = 10^y

{{{x = 10^1}}}

So, x = 10

Check:
{{{7^(logx) = 7}}}
{{{7^log10 = 7}}} but log10 = 1
{{{7^1 = 7}}}