Answer
$3.46$ %
Work Step by Step
Let $f(x)$ = $\ln(1+x)$, $a=0$
$f(a)$ = $f(0)$ = $0$
$f'(x)$ = $\frac{1}{1+x}$
$f'(a)$ = $f'(0)$ = $1$
$L(x)$ = $f'(a)(x-a)+f(a)$
$L(x)$ = $1(x-0)+0$ = $x$
Percentage error = $|\frac{\ln1.07-1}{\ln1.07}|\times100$ = $3.46$ %