Dry food is often cheaper and easier to maintain than wet food. Dry food allows you to leave the food out for the day without having to worry about it going bad. Most adult cats can thrive on dry cat food that is high quality. Make sure the dry cat food contains mostly animal proteins. Your cat may prefer dry cat food over wet food, especially if you introduce dry food to the cat early in its development. Wet cat food can be more expensive and takes a bit more effort to maintain. It can be a good option if your cat is prone to dehydration and urinary tract issues, and is great for maintaining weight since it provides less calories for the same volume. Wet cat food is perishable so it should be given to your cat at feeding time as soon as you open it. You should then dispose of any remaining wet cat food so it does not go bad. Rotating your cat’s food can help them stay healthy and encourage them to eat at every meal time. Try giving your cat a mix of wet and dry food, alternating between wet and dry. You can also give the cat wet food 1 or 2 times a week and dry food the rest of the week, especially if you are on a budget. You can also try giving your cat different brands of dry and wet food to see which one it likes best.
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One-sentence summary -- Pick dry food for convenience. Feed the cat wet food if it tends to get dehydrated. Give the cat a mix of wet and dry food.


Tulip-shaped glasses are a favorite among scotch drinkers. The tulip shape concentrates the scotch’s aromas at your nose, making them easier to detect. If you want a more classic look to your scotch glass, use a tumbler glass. Hold the neck of the bottle about 1 inch (2.5 cm) above your glass, taking care not to spill any scotch. Pour about 1.5 fluid ounces (44 mL), or 1 shot-glass full, of scotch into your glass. Water doesn’t dilute scotch; it actually opens up the flavor and releases more of the scotch’s aromas. Use a splash of water to take the edge off your dram if it’s over 40% alcohol (80-proof). Ice dulls the flavors in scotch and conceals some of the pleasant aromas. It can also numb your taste buds, which isn't ideal if you're trying to learn about scotch. However, you can add ice if you want. Or, keep your dram cold using frozen whiskey stones instead.
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One-sentence summary -- Find a glass with a tapered mouth. Pour yourself a dram. Add water to the scotch if it’s high in alcohol. Note that adding ice dilutes the flavor.


This method covers problems in the form logb⁡(x)logb⁡(a){\displaystyle {\frac {\log _{b}(x)}{\log _{b}(a)}}}. However, it does not work for a few special cases:  The log of a negative number is undefined for all bases (such as log⁡(−3){\displaystyle \log(-3)} or log4⁡(−5){\displaystyle \log _{4}(-5)}). Write "no solution." The log of zero is also undefined for all bases. If you see a term such as ln⁡(0){\displaystyle \ln(0)}, write "no solution." The log of one in any base (log⁡(1){\displaystyle \log(1)}) always equals zero, since x0=1{\displaystyle x^{0}=1} for all values of x. Replace that logarithm with 1 instead of using the method below. If the two logarithms have different bases, such as log3(x)log4(a){\displaystyle {\frac {log_{3}(x)}{log_{4}(a)}}}, and you cannot simplify either one into an integer, the problem is not feasible to solve by hand. Assuming you did not find any of the exceptions above, you can now simplify the problem into one logarithm. To do this, use the formula logb⁡(x)logb⁡(a)=loga⁡(x){\displaystyle {\frac {\log _{b}(x)}{\log _{b}(a)}}=\log _{a}(x)}.  Example 1: Solve the problem log⁡16log⁡2{\displaystyle {\frac {\log {16}}{\log {2}}}}.Start by converting this into one logarithm using the formula above: log⁡16log⁡2=log2⁡(16){\displaystyle {\frac {\log {16}}{\log {2}}}=\log _{2}(16)}. This formula is the "change of base" formula, derived from basic logarithmic properties. Remember, to solve loga⁡(x){\displaystyle \log _{a}(x)}, think "a?=x{\displaystyle a^{?}=x}" or "What exponent can I raise a by to get x?" It's not always feasible to solve this without a calculator, but if you're lucky, you'll end up with an easily simplified logarithm. Example 1 (cont.): Rewrite log2⁡(16){\displaystyle \log _{2}(16)} as 2?=16{\displaystyle 2^{?}=16}. The value of "?" is the answer to the problem. You may need to find it by trial and error:22=2∗2=4{\displaystyle 2^{2}=2*2=4}23=4∗2=8{\displaystyle 2^{3}=4*2=8}24=8∗2=16{\displaystyle 2^{4}=8*2=16}16 is what you were looking for, so log2⁡(16){\displaystyle \log _{2}(16)} = 4. Some logarithms are very difficult to solve by hand. You'll need a calculator if you need the answer for a practical purpose. If you're solving problems in math class, your teacher most likely expects you to leave the answer as a logarithm. Here's another example using this method on a more difficult problem:  Example 2: What is log3⁡(58)log3⁡(7){\displaystyle {\frac {\log _{3}(58)}{\log _{3}(7)}}}? Convert this into one logarithm: log3⁡(58)log3⁡(7)=log7⁡(58){\displaystyle {\frac {\log _{3}(58)}{\log _{3}(7)}}=\log _{7}(58)}. (Notice that the 3 in each initial log disappears; this is true for any base.) Rewrite as 7?=58{\displaystyle 7^{?}=58} and test possible values of ?:72=7∗7=49{\displaystyle 7^{2}=7*7=49}73=49∗7=343{\displaystyle 7^{3}=49*7=343}Since 58 falls between these two numbers, log7⁡(58){\displaystyle \log _{7}(58)} has no integer answer. Leave your answer as log7⁡(58){\displaystyle \log _{7}(58)}.
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One-sentence summary --
Check for negative numbers and ones. Convert the expression into one logarithm. Calculate by hand if possible. Leave the answer in logarithm form if you cannot simplify it.