Question

If $f(x+y)=f\left(x\right)f\left(y\right)$ and $f\left(5\right)=32$ then $f\left(7\right)=$

• A. $35$
• B. $36$
• C. $\frac{7}{5}$
• D. $128$

Answer 1

The correct option is D $128$

$f(x+y)=f\left(x\right).f\left(y\right)$
Put $x=1,y=1$
$f(x+y)=f\left(x\right).f\left(y\right)$
$\Rightarrow f\left(2\right)=f\left(1\right)f\left(1\right)$
$\Rightarrow f\left(3\right)=f\left(2\right)f\left(1\right)=f(1{)}^3$
$\Rightarrow f\left(4\right)=f\left(3\right)f\left(1\right)=f(1{)}^4$
$\Rightarrow f\left(5\right)=f\left(4\right)f\left(1\right)=f(1{)}^5$
Given$f\left(5\right)=32$

$\therefore f\left(1\right)=2$

And $f\left(2\right)=4$
Now,
$f\left(7\right)=f\left(5\right)f\left(2\right)=32\times 4=128$