First, because we have a quotient of two expressions, we can use the quotient rule:

[latex]\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right)=\mathrm{ln}\left({x}^{4}y\right)-\mathrm{ln}\left(7\right)[/latex]

Then seeing the product in the first term, we use the product rule:

[latex]\mathrm{ln}\left({x}^{4}y\right)-\mathrm{ln}\left(7\right)=\mathrm{ln}\left({x}^{4}\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)[/latex]

Finally, we use the power rule on the first term:

[latex]\mathrm{ln}\left({x}^{4}\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)=4\mathrm{ln}\left(x\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)[/latex]

[latex]\begin{array}{l}\mathrm{log}\left(\sqrt{x}\right)\hfill & =\mathrm{log}{x}^{\left(\frac{1}{2}\right)}\hfill \\ \hfill & =\frac{1}{2}\mathrm{log}x\hfill \end{array}[/latex]

We can expand by applying the product and quotient rules.

[latex]\begin{array}{lllllll}{\mathrm{log}}_{6}\left(\frac{64{x}^{3}\left(4x+1\right)}{\left(2x - 1\right)}\right)\hfill & ={\mathrm{log}}_{6}64+{\mathrm{log}}_{6}{x}^{3}+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & \text{Apply the product and quotient rule}.\hfill \\ \hfill & ={\mathrm{log}}_{6}{2}^{6}+{\mathrm{log}}_{6}{x}^{3}+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & {\text{Simplify by writing 64 as 2}}^{6}.\hfill \\ \hfill & =6{\mathrm{log}}_{6}2+3{\mathrm{log}}_{6}x+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & \text{Apply the power rule}.\hfill \end{array}[/latex]

Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression [latex]4\mathrm{ln}\left(x\right)[/latex], we identify the factor, 4, as the exponent and the argument, x, as the base and rewrite the product as a logarithm of a power:

[latex]4\mathrm{ln}\left(x\right)=\mathrm{ln}\left({x}^{4}\right)[/latex].

From left to right, since we have the addition of two logs, we first use the product rule:

[latex]{\mathrm{log}}_{3}\left(5\right)+{\mathrm{log}}_{3}\left(8\right)={\mathrm{log}}_{3}\left(5\cdot 8\right)={\mathrm{log}}_{3}\left(40\right)[/latex]

This simplifies our original expression to:

[latex]{\mathrm{log}}_{3}\left(40\right)-{\mathrm{log}}_{3}\left(2\right)[/latex]

Using the quotient rule:

[latex]{\mathrm{log}}_{3}\left(40\right)-{\mathrm{log}}_{3}\left(2\right)={\mathrm{log}}_{3}\left(\frac{40}{2}\right)={\mathrm{log}}_{3}\left(20\right)[/latex]

We apply the power rule first:

[latex]{\mathrm{log}}_{2}\left({x}^{2}\right)+\frac{1}{2}{\mathrm{log}}_{2}\left(x - 1\right)-3{\mathrm{log}}_{2}\left({\left(x+3\right)}^{2}\right)={\mathrm{log}}_{2}\left({x}^{2}\right)+{\mathrm{log}}_{2}\left(\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)[/latex]

From left to right, since we have the addition of two logs, we apply the product rule to the sum:

[latex]{\mathrm{log}}_{2}\left({x}^{2}\right)+{\mathrm{log}}_{2}\left(\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)={\mathrm{log}}_{2}\left({x}^{2}\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)[/latex]

Finally we apply the quotient rule to the difference:

[latex]{\mathrm{log}}_{2}\left({x}^{2}\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)={\mathrm{log}}_{2}\frac{{x}^{2}\sqrt{x - 1}}{{\left(x+3\right)}^{6}}[/latex]

We apply the power rule first:

[latex]2\mathrm{log}x - 4\mathrm{log}\left(x+5\right)+\frac{1}{x}\mathrm{log}\left(3x+5\right)=\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}\right)+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)[/latex]

From left to right, since we have the difference of two logs, we apply the quotient rule to the difference:

[latex]\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}\right)+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)=\mathrm{log}\left(\frac{{x}^{2}}{{\left(x+5\right)}^{4}}\right)+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)[/latex]

Finally we apply the product rule to the sum:

[latex]\mathrm{log}\left(\frac{{x}^{2}}{{\left(x+5\right)}^{4}}\right)+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)=\mathrm{log}\left(\frac{{{x}^{2}}{{\left(3x+5\right)}^{{x}^{-1}}}}{{\left(x+5\right)}^{4}}\right)[/latex]

Suppose C is the original concentration of hydrogen ions and P is the original pH of the liquid. Then [latex]\text{P}=-\mathrm{log}\left(C\right)[/latex]. If the concentration is doubled, the new concentration is 2C. Then the pH of the new liquid is [latex]\text{pH}=-\mathrm{log}\left(2C\right)[/latex]

Using the product rule of logs

[latex]\text{pH}=-\mathrm{log}\left(2C\right)=-\left(\mathrm{log}\left(2\right)+\mathrm{log}\left(C\right)\right)=-\mathrm{log}\left(2\right)-\mathrm{log}\left(C\right)[/latex]

Since [latex]P=-\mathrm{log}\left(C\right)[/latex], the new pH is

[latex]\text{pH}=P-\mathrm{log}\left(2\right)\approx P - 0.301[/latex]

When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.