## College Algebra (10th Edition)

$x=2 - \log{2.5}$
Divide by 2 on both sides of the equation to obtain: $\dfrac{2 \cdot 10^{2-x}}{2} = \dfrac{5}{2} \\10^{2-x}=2.5$ Take the common logarithm of both sides to obtain: $\log{(10^{2-x})}=\log{2.5}$ Note that $\log{(10^x)} = x$. Thus, the equation above is equivalent to: $2-x=\log{2.5}$ Subtract by $2$ on both sides of the equation to obtain: $2-x-2 = \log{2.5} - 2 \\-x = \log{2.5} - 2$ Multiply by $-1$ on both sides of the equation to obtain: $-1(-x) = -1(\log{2.5}-2) \\x = -\log{2.5} + 2 \\x=2 - \log{2.5}$