Solve for x 2^(x-1)<3

$2^{x - 1} < 3$

Step 1

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

$\ln (2^{x - 1}) < \ln (3)$

Step 2

Expand $\ln (2^{x - 1})$ by moving $x - 1$ outside the logarithm.

$(x - 1) \ln (2) < \ln (3)$

Step 3

Simplify the left side.

Simplify $(x - 1) \ln (2)$ .

Apply the distributive property.

$x \ln (2) - 1 \ln (2) < \ln (3)$

Rewrite $- 1 \ln (2)$ as $- \ln (2)$ .

$x \ln (2) - \ln (2) < \ln (3)$

Step 4

Move all the terms containing a logarithm to the left side of the equation.

$x \ln (2) - \ln (2) - \ln (3) = 0$

Step 5

Move all terms not containing $x$ to the right side of the equation.

Add $\ln (2)$ to both sides of the equation.

$x \ln (2) - \ln (3) = \ln (2)$

Add $\ln (3)$ to both sides of the equation.

$x \ln (2) = \ln (2) + \ln (3)$

Step 6

Divide each term in $x \ln (2) = \ln (2) + \ln (3)$ by $\ln (2)$ and simplify.

Divide each term in $x \ln (2) = \ln (2) + \ln (3)$ by $\ln (2)$ .

$\frac{x \ln (2)}{\ln (2)} = \frac{\ln (2)}{\ln (2)} + \frac{\ln (3)}{\ln (2)}$

Simplify the left side.

Cancel the common factor of $\ln (2)$ .

Cancel the common factor.

$\frac{x \ln (2)}{\ln (2)} = \frac{\ln (2)}{\ln (2)} + \frac{\ln (3)}{\ln (2)}$

Divide $x$ by $1$ .

$x = \frac{\ln (2)}{\ln (2)} + \frac{\ln (3)}{\ln (2)}$

Simplify the right side.

Cancel the common factor of $\ln (2)$ .

Cancel the common factor.

$x = \frac{\ln (2)}{\ln (2)} + \frac{\ln (3)}{\ln (2)}$

Rewrite the expression.

$x = 1 + \frac{\ln (3)}{\ln (2)}$

Step 7

The solution consists of all of the true intervals.

$x < 1 + \frac{\ln (3)}{\ln (2)}$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$x < 1 + \frac{\ln (3)}{\ln (2)}$

Interval Notation:

$(- \infty, 1 + \frac{\ln (3)}{\ln (2)})$

Step 9

Solve for x 2^(x-1)<3

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