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Question Number 93077 by hmamarques1994@gmail.com last updated on 10/May/20

$$\mathrm{Se}\mathrm{f}(\sqrt{\mathrm{x}}-1)=\mathrm{x}+6,\log \left[\mathrm{f}\left(1\right)\right]=?$$

Commented by john santu last updated on 11/May/20

$$\mathrm{f}(\sqrt{x}-1)=x+6$$$$set\sqrt{x}-1=p\Rightarrow x={(p+1)}^2$$$$sof\left(x\right)={(x+1)}^2+6$$$$f\left(1\right)=10$$$$\log {}_{10}f\left(1\right)=\log {}_{10}\left(10\right)=1$$$$\mathrm{if}\mathrm{you}\mathrm{meant}\log \mathrm{f}\left(1\right)=\ln \mathrm{f}\left(1\right)=\ln \left(10\right)$$$$$$

Answered by Kunal12588 last updated on 10/May/20

$$\sqrt{x}-1=1$$$$\Rightarrow x=4$$$${log}_{10}\left(f\left(1\right)\right)=1$$

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