Rewrite the base and argument of the logarithm with a common base.

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4) $4 - \log_{25} \frac{1}{125}$ can be simplified to

Rewrite the base and argument of the logarithm with a common base.
$25 = 5^2$
$\frac{1}{125} = \frac{1}{5^3} = 5^{-3}$
Substitute these into the logarithm.
$\log_{25} \frac{1}{125} = \log_{5^2} 5^{-3}$
Use the logarithm property $\log_{b^k} a^m = \frac{m}{k} \log_b a$ .
$\log_{5^2} 5^{-3} = \frac{-3}{2} \log_5 5$
Since $\log_5 5 = 1$ .
$\frac{-3}{2} \times 1 = -\frac{3}{2}$
Substitute this value back into the original expression.
$4 - \left(-\frac{3}{2}\right) = 4 + \frac{3}{2}$
Add the fractions.
$4 + \frac{3}{2} = \frac{8}{2} + \frac{3}{2} = \frac{11}{2}$
The simplified expression is $\boxed{\frac{11}{2}}$ .

5) expanding $(2x-1)^2$ gives
Use the algebraic identity $(a-b)^2 = a^2 - 2ab + b^2$ .
Here, $a = 2x$ and $b = 1$ .
Substitute $a$ and $b$ into the identity.
$(2x-1)^2 = (2x)^2 - 2(2x)(1) + (1)^2$
Simplify each term.
$= 4x^2 - 4x + 1$
Expanding $(2x-1)^2$ gives $\boxed{4x^2 - 4x + 1}$ .

7) The set of value of $x$ for which $2|x-5| < 1$ is
Isolate the absolute value term.
$2|x-5| < 1$ $|x-5| < \frac{1}{2}$
Convert the absolute value inequality into a compound inequality.
For $|u| < c$ , it means $-c < u < c$ .
$-\frac{1}{2} < x-5 < \frac{1}{2}$
Add 5 to all parts of the inequality to solve for $x$ .
$-\frac{1}{2} + 5 < x < \frac{1}{2} + 5$ $-\frac{1}{2} + \frac{10}{2} < x < \frac{1}{2} + \frac{10}{2}$ $\frac{9}{2} < x < \frac{11}{2}$
The set of values for $x$ is $\boxed{\left(\frac{9}{2}, \frac{11}{2}\right)}$ .

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