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Question Number 213203 by golsendro last updated on 01/Nov/24

$$\cancel{\underset{―}{\underbrace{\mathcal{G}}}}$$

Answered by lepuissantcedricjunior last updated on 01/Nov/24

$$\mathit{f}\left(\mathit{x}\mathit{y}\right)=\mathit{f}\left(\mathit{x}\right)+\mathit{f}\left(\mathit{y}\right)$$$$\mathit{f}\left(10\right)=14;\mathit{f}\left(20\right)=40$$$$\mathit{c}\mathit{a}\mathit{l}\mathit{c}\mathit{u}\mathit{l}\mathit{o}\mathit{n}\mathit{s}\mathit{f}\left(500\right)$$$$\mathit{o}\mathit{n}\mathit{a}\mathit{f}\left(500\right)=\mathit{f}\left(50\times 10\right)=\mathit{f}\left(50\right)+\mathit{f}\left(10\right)$$$$=\mathit{f}\left(5\right)+2\mathit{f}\left(10\right)$$$$\mathit{o}\mathit{r}\mathit{f}\left(20\right)=\mathit{f}\left(10\right)+\mathit{f}\left(2\right)=>\mathit{f}\left(2\right)=26$$$$\mathit{f}\left(10\right)=\mathit{f}\left(5\right)+\mathit{f}\left(2\right)=>\mathit{f}\left(5\right)=-12$$$$=>\mathit{f}\left(500\right)=-12+28=16$$

Answered by mr W last updated on 01/Nov/24

$$f\left(20\right)=f\left(2\times 10\right)=f\left(2\right)+f\left(10\right)$$$$\Rightarrow f\left(2\right)=40-14=26$$$$f\left(x^n\right)=nf\left(x\right)$$$$f\left(500\right)=f\left(\frac{1000}{2}\right)=f\left({10}^3\times 2^{-1}\right)$$$$=3f\left(10\right)-f\left(2\right)$$$$=3\times 14-26=16✓$$

Answered by cherokeesay last updated on 01/Nov/24

Answered by Rasheed.Sindhi last updated on 02/Nov/24

$$f\left(500\right)=f\left(2^2\times 5^3\right)=2f\left(2\right)+3f\left(5\right)...A$$$$Soweneedf\left(2\right)andf\left(5\right)forf\left(500\right)$$$$$$$$f\left(10\right)=f\left(2\times 5\right)=f\left(2\right)+f\left(5\right)=14...\left(i\right)$$$$f\left(20\right)=f\left(2^2\times 5\right)=2f\left(2\right)+f\left(5\right)=40...\left(ii\right)$$$$\left(ii\right)-\left(i\right):f\left(2\right)=26$$$$\left(i\right)\Rightarrow f\left(5\right)=14-26=-12$$$$A\Rightarrow f\left(500\right)=2\left(26\right)+3(-12)=16$$

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