Solve for x log base 0.5 of 15=1-2x

$\log_{0.5} (15) = 1 - 2 x$

Step 1

Rewrite the equation as $1 - 2 x = \log_{0.5} (15)$ .

$1 - 2 x = \log_{0.5} (15)$

Step 2

Subtract $1$ from both sides of the equation.

$- 2 x = \log_{0.5} (15) - 1$

Step 3

Divide each term in $- 2 x = \log_{0.5} (15) - 1$ by $- 2$ and simplify.

Divide each term in $- 2 x = \log_{0.5} (15) - 1$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{\log_{0.5} (15)}{- 2} + \frac{- 1}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{\log_{0.5} (15)}{- 2} + \frac{- 1}{- 2}$

Divide $x$ by $1$ .

$x = \frac{\log_{0.5} (15)}{- 2} + \frac{- 1}{- 2}$

Simplify the right side.

Simplify each term.

Rewrite $\frac{\log_{0.5} (15)}{- 2}$ as $\frac{1}{- 2} \log_{0.5} (15)$ .

$x = \frac{1}{- 2} \log_{0.5} (15) + \frac{- 1}{- 2}$

Simplify $\frac{1}{- 2} \log_{0.5} (15)$ by moving $- \frac{1}{- 2}$ inside the logarithm.

$x = - \log_{0.5} (15^{- \frac{1}{- 2}}) + \frac{- 1}{- 2}$

Move the negative in front of the fraction.

$x = - \log_{0.5} (15^{- - \frac{1}{2}}) + \frac{- 1}{- 2}$

Multiply $- - \frac{1}{2}$ .

Multiply $- 1$ by $- 1$ .

$x = - \log_{0.5} (15^{1 (\frac{1}{2})}) + \frac{- 1}{- 2}$

Multiply $\frac{1}{2}$ by $1$ .

$x = - \log_{0.5} (15^{\frac{1}{2}}) + \frac{- 1}{- 2}$

Dividing two negative values results in a positive value.

$x = - \log_{0.5} (15^{\frac{1}{2}}) + \frac{1}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = - \log_{0.5} (15^{\frac{1}{2}}) + \frac{1}{2}$

Decimal Form:

$x = 2.45344529 \dots$