Problem 77 - Algebra and Trigonometry

Question

Solve: \(\log _{5}(x+3)=2\)

Step-by-Step Solution

Answer

x = 22

Step 1: Understand the Problem

The given equation is \(\text{log}_{5}(x+3) = 2\). We need to solve for \(x\). This is a logarithmic equation with base 5.

Step 2: Rewrite the Equation in Exponential Form

To solve the logarithmic equation, rewrite it in its exponential form. The equation \(\text{log}_{5}(x+3) = 2\) can be rewritten as \({5^2 = x + 3}\).

Step 3: Calculate the Exponent

Compute \({5^2}\). \({5^2}\) is 25.

Step 4: Solve for x

Now, we have the equation \(25 = x + 3\). To solve for \(x\), subtract 3 from both sides: \(x = 25 - 3\).

Step 5: Simplify

Simplify the equation to find \(x = 22\).

Key Concepts

logarithms

Logarithms are a way to represent exponential relationships. They answer the question: 'To what power must the base be raised, to produce a given number?' For instance, in our problem, \(\text{log}_5(x+3)=2\), it reads as 'To what power must 5 be raised, to get \((x+3)\)?' Understanding how to translate between logarithms and exponential forms is key. Here, \(5^2\) equals 25, letting us transition from logarithmic to exponential format seamlessly.

exponential functions

Exponential functions are equations where variables appear as exponents. They grow rapidly and are essential in many areas of math and science. In our exercise, rewriting \(\text{log}_5(x+3) = 2\) into its exponential form, we convert it to \(5^2 = x + 3\). This step helps simplify the problem as it is easier to manage a direct calculation involving exponents.

step-by-step solutions

Breaking down problems into smaller, manageable steps can simplify complex equations. In this exercise, we gradually convert and solve the logarithmic equation:

• Starting with \(\text{log}_5(x+3) = 2\), we translate it to its exponential form \(5^2 = x + 3\).
• Then, we calculate the power \(5^2 = 25\).
• Finally, we solve for \((x\), getting \(x + 3 = 25\) and isolating \(x\) by subtracting 3 from 25, resulting in \(x = 22\).

This methodical approach helps prevent mistakes and ensures thorough understanding.

algebra

Algebra forms the foundation of solving equations. It’s all about finding the unknown values by performing various operations on known values. For our logarithmic equation, we applied algebraic principles to isolate \(x\). Starting with \(25 = x + 3\), we simply subtracted 3 from both sides to solve for \(x\). Such operations are fundamental in algebra, allowing us to manipulate equations and solve for unknowns effectively.