Solve for x log base square root of 2 of x+ log base square root of 2 of x+1=2

$\log_{\sqrt{2}} (x) + \log_{\sqrt{2}} (x + 1) = 2$

Simplify the left side.

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{\sqrt{2}} (x (x + 1)) = 2$

Logarithm base $\sqrt{2}$ of $x (x + 1)$ is $N a N$ .

Rewrite as an equation.

$\log_{\sqrt{2}} (x (x + 1)) = x$

Rewrite $\log_{\sqrt{2}} (x (x + 1)) = x$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b$ does not equal $1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

${\sqrt{2}}^{x} = x (x + 1)$

Create equivalent expressions in the equation that all have equal bases.

${\sqrt{2}}^{x} = {\sqrt{2}}^{N a N}$

Since the bases are the same, the two expressions are only equal if the exponents are also equal.

$x = N a N$

The variable $x$ is equal to $N a N$ .

$N a N = 2$

Multiply $N$ by $N$ by adding the exponents.

Move $N$ .

$N \cdot N a = 2$

Multiply $N$ by $N$ .

$N^{2} a = 2$

The variable $x$ got canceled for any value of $x$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

Solve for x log base square root of 2 of x+ log base square root of 2 of x+1=2

Related Questions

Support LCMGCF.com by disabling your adblocker.