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Chapter 4: Problem 7

Simplify each expression. $$\log \left(10^{y+1}\right)$$

Short Answer

Expert verified
y + 1

Step by step solution

01

Identify the logarithm property

Recognize and identify the logarithm property that will help in simplifying the expression. The relevant property is the power rule for logarithms: \ \ \ \ \ \ \ \ \ \text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{ log }} \text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{a}} } } }}} } }}}}}}}}}}}}}}}}}}}}}}}}}}}$$} $$ $$ \( \)} $$ } \( } {x} = y x }} }} $$ ) }} \)} $$}\( $$}\)} \( }} }} \)} $$}\(} \) }} }} ${

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithm Simplification
Logarithm simplification makes expressions easier to understand and compute. One of the key properties that help simplify logs is the power rule, which will be discussed in detail later. Simplifying logarithms often involves using well-established rules and properties.

For example, consider the expression \(\log(10^{y+1})\). To simplify this, we recognize that 10 is the base of our logarithm (common logarithm). We can use the power rule to simplify this expression, which states \(\log_b(a^c) = c \, \log_b(a)\). We'll dig deeper into this rule in the next section.
Power Rule for Logarithms
The power rule for logarithms is a very helpful property that simplifies many logarithmic expressions.

The rule states: \(\log_b(a^c) = c \, \log_b(a)\). This means that the exponent \(c\) can be brought down in front of the logarithm as a multiplier.

Let's apply the power rule to our earlier example:
  • Starting Expression: \(\log(10^{y+1})\)
  • Apply Power Rule: \(\log(10^{y+1}) = (y+1) \, \log(10)\)
  • Since \(\log(10) = 1\) for any common logarithm (base 10), our expression simplifies to: \(y + 1\)
This property simplifies the calculation and helps to better understand logarithmic equations.
Logarithmic Equations
Logarithmic equations are equations that involve logarithms. Solving these types of equations often means utilizing several logarithm properties, like the power rule.

When given a logarithmic equation, our goal is to isolate the variable. Sometimes, we may need to simplify the logarithm first.

For instance, solving \(\log(10^{y+1}) = 2\) can be broken down into:
  • First, simplify using the power rule: \(\log(10^{y+1}) = y+1\)
  • Set the equation: \(y+1 = 2\)
  • Solve for \(y\): \(y = 1\)
With these steps, we isolate the variable and find the solution. By simplifying logarithms and using properties like the power rule, logarithmic equations become easier to solve.

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Most popular questions from this chapter

Solve each problem. Find the annual percentage rate compounded continuously to the nearest tenth of a percent for which \(\$ 10\) would grow to \(\$ 30\) for each of the following time periods. a. 5 years b. 10 years c. 20 years d. 40 years

Use the following definition. In chemistry, the \(\mathrm{pH}\) of a solution is defined to be $$\mathrm{pH}=-\log \left[H^{+}\right],$$ where \(H^{+}\) is the hydrogen ion concentration of the solution in moles per liter. Distilled water has a pH of approximately 7. A substance with a pH under 7 is called an acid, and one with a pH over 7 is called a base. Tomato juice has a hydrogen ion concentration of \(10^{-4.1}\) moles per liter. Find the \(\mathrm{pH}\) of tomato juice.

Solve each problem. When needed, use 365 days per year and 30 days per month. Periodic Compounding Melinda invests her \(\$ 80,000\) winnings from Publishers Clearing House at a \(9 \%\) annual percentage rate. Find the amount of the investment at the end of 20 years and the amount of interest earned during the 20 years if the interest is compounded a. annually b. quarterly c. monthly d. daily.

Solve each equation. Find the exact solutions. $$3^{x+1}=7$$

Calculators that perform exponential regression often use \(y=a \cdot b^{x}\) as the exponential growth model instead of \(y=a \cdot e^{c x} .\) For what value of \(c\) is \(a \cdot b^{x}=a \cdot e^{c x} ?\) If a calculator gives \(y=500(1.036)^{x}\) for a growth model, then what is the continuous growth rate to the nearest hundredth of a percent?
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