Linear Algebra Examples

\log_{2} (x + 5) - \log ({(2)}^{x}) = 3

$\log_{2} (x + 5) - \log ({(2)}^{x}) = 3$

Step 1

Set the argument in

\log_{2} (x + 5)

$\log_{2} (x + 5)$ greater than

0

$0$ to find where the expression is defined.

x + 5 > 0

$x + 5 > 0$

Step 2

Subtract

5

$5$ from both sides of the inequality.

x > - 5

$x > - 5$

Step 3

Set the argument in

\log ({(2)}^{x})

$\log ({(2)}^{x})$ greater than

0

$0$ to find where the expression is defined.

{(2)}^{x} > 0

${(2)}^{x} > 0$

Step 4

Solve for

x

$x$ .

Tap for more steps...

Step 4.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

\ln (2^{x}) > \ln (0)

$\ln (2^{x}) > \ln (0)$

Step 4.2

The equation cannot be solved because

\ln (0)

$\ln (0)$ is undefined.

Undefined

Step 4.3

There is no solution for

2^{x} > 0

$2^{x} > 0$

No solution

Step 5

The domain is all values of

x

$x$ that make the expression defined.

Interval Notation:

(- 5, \infty)

$(- 5, \infty)$

Set-Builder Notation:

{x | x > - 5}

${x | x > - 5}$

Step 6