Question

If log 2 = 0 . 30103, log 3 = 0 . 4771213, log 7 = 0 . 8450980, then solve :
$3^{1+x}=7^{x/2}$

Answer 1

We write the given equation as $(7\times 3{)}^x=2^{2x+1}.(10/2{)}^x$
Taking logarithm of both side, we have
x( log 7 + log 3) = (2x + 1 ) log + 2 x(log 10 - log 2)
or x[log 7 + log 3 - 2 log 2 - log 10 + log 2] = 2
or x[log 7 + log 3 - log 2 - 1] = log 2
or x[0.8450980 + 0.4771213 - 0.3010300 - 1] = 0.03013
or x(0.0211893) = 0.30103
$\therefore$ x = 14.207 (approx).