$Solve for x in the equation$ $l o g [l o g (2 + l o g_{2} (x + 1))] = 0$

Open in App

Solution

Rewrite the equation as
$l o g [l o g (2 + l o g_{2} (x + 1))] = l o g (1)$ ,
$[since log(1) = 0]$

$l o g (2 + l o g_{2} (x + 1)) = 1$

$2 + l o g_{2} (x + 1) = 10$

$l o g_{2} (x + 1) = 8$

$x + 1 = 2^{8}$

$x = 2^{8} - 1$

Suggest Corrections

Similar questions

x > 0

l o g_{2} x + l o g_{2} (\sqrt{x}) + l o g_{2} (\sqrt[4]{x}) + l o g_{2} (\sqrt[8]{x}) + . . . \infty = 4,

x =

l o g_{2} x + l o g_{2} (\sqrt{x}) + l o g_{2} (\sqrt[4]{x}) + l o g_{2} (\sqrt[8]{x}) + l o g_{2} (\sqrt[16]{x}) + . . . . . = 4

$Solve for x in the equation$ $l o g [l o g (2 + l o g_{2} (x + 1))] = 0$

log 2, log (2^{x} - 1), log (2^{x} + 3)

x

Join BYJU'S Learning Program