Se-f-x-1-x-6-log-f-1-

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

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

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

f (\sqrt{x} - 1) = x + 6

s e t \sqrt{x} - 1 = p \Rightarrow x = {(p + 1)}^{2}

s o f (x) = {(x + 1)}^{2} + 6

f (1) = 10

\log_{10} f (1) = \log_{10} (10) = 1

if you meant \log f (1) = \ln f (1) = \ln (10)

Answered by Kunal12588 last updated on 10/May/20

\sqrt{x} - 1 = 1

\Rightarrow x = 4

{l o g}_{10} (f (1)) = 1