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Chapter 15: Problem 139

Without using a calculator, what is the base- 10 logarithm of \(10^{0}\) ? Of 1?

Short Answer

Expert verified
The base-10 logarithm of both \(10^0\) and 1 is 0.

Step by step solution

01

Understand the properties of logarithms

Logarithms have a few important properties, but for this problem, we will mainly focus on the property that states: \(\log_a(a^n) = n\). In other words, if you take the logarithm with a base of some number 'a' of that same number raised to the power 'n', you will simply get 'n' back.
02

Calculate the base-10 logarithm of \(10^0\)

In this case, our base is 10 and our exponent, n, is 0. We can apply the property mentioned in step 1 to get the value of the base-10 logarithm. \(\log_{10}(10^0) = 0\) So the base-10 logarithm of \(10^0\) is 0.
03

Calculate the base-10 logarithm of 1

Now we are asked to find the logarithm value for the number 1. We can think about what number n, the exponent, we can raise 10 to in order to get the value of 1. \(10^n = 1\) With the knowledge that any non-zero number raised to the power of 0 is 1, we can deduce that the exponent here is 0 (since \(10^0 = 1\)). Now, using the property from step 1, we can say: \(\log_{10}(1) = \log_{10}(10^0) = 0\) So the base-10 logarithm of 1 is also 0. In conclusion, the base-10 logarithm of both \(10^0\) and 1 is 0.

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Most popular questions from this chapter

Hydride ion, \(\mathrm{H}\), is an exceptionally strong base, reacting with water to produce lots of hydroxide ion and \(\mathrm{H}_{2}\) gas. The \(K_{\mathrm{eq}}\) for this reaction is huge. (a) Write the balanced equation for the reaction between hydride and water. (b) Explain why \(\mathrm{H}_{2}\) gas forms. (Hint: Use the Bronsted-Lowry definition of base.)

A solution is prepared by dissolving \(2.00\) moles of \(\mathrm{HNO}_{3}\) in enough water to get \(800.0 \mathrm{~mL}\) of solution. What are the \(\mathrm{H}_{3} \mathrm{O}^{+}\) and the \(\mathrm{OH}\) molar concentrations?

How many moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) would it take to completely neutralize \(0.4\) mole of phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4} ?\)

Consider the following statement: "As the \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration in an aqueous solution increases, the \(\mathrm{OH}^{-}\) concentration must decrease." (a) Why is this true? (b) Is there any way to stop the \(\mathrm{OH}^{-}\) concentration from decreasing as you increase the \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration?

Write a chemical equation for the reaction between each pair of reactants, using single or double arrows as appropriate: (a) \(\mathrm{HNO}_{3}\) and \(\mathrm{OH}^{-}\) (b) \(\mathrm{HF}\) and \(\mathrm{OH}^{-}\) (c) \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{HCO}_{3}^{-}\) and \(\mathrm{H}_{2} \mathrm{O}\)
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