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91Ó°ÊÓ

Solve the equation. $$6^{7-x}=6^{2 x+1}$$

Short Answer

Expert verified
The solution is \(x = 2\).

Step by step solution

01

Understanding the Equation

The given equation is \(6^{7-x} = 6^{2x+1}\). Since the bases on both sides of the equation are the same, we can set the exponents equal to each other to solve for \(x\).
02

Setting Exponents Equal

Since \(6^{a} = 6^{b}\) implies \(a = b\), we equate the exponents: \[7 - x = 2x + 1\]
03

Rearranging the Equation

Rearrange the terms to isolate \(x\) on one side of the equation. Start by adding \(x\) to both sides:\[7 = 3x + 1\].
04

Solving for x

Subtract 1 from both sides to simplify:\[6 = 3x\].
05

Isolating x

Divide both sides of the equation by 3:\[x = 2\].

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Key Concepts

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

Solving Equations
To solve equations, the main goal is to find the value of the unknown variable that makes the equation true. In our example, we're dealing with the equation \(6^{7-x} = 6^{2x+1}\).

The process begins with recognizing that when you have the same base on both sides of an exponential equation, you can set the exponents equal to each other. This is possible because if \(a^m = a^n\), then \(m = n\).

By equating the exponents \((7-x) = (2x + 1)\), we transform an exponential equation into a simpler linear equation. We then solve this linear equation through a series of algebraic manipulations to find the value of \(x\). This method simplifies the process considerably and avoids unnecessary complications.
Exponent Rules
Exponent rules are essential in algebra and help simplify expressions involving powers. In the equation \(6^{7-x} = 6^{2x+1}\), we make use of the rule that allows us to equate exponents when the bases are identical.

This rule states that if \(a^b = a^c\), then \(b = c\). This property is incredibly useful because it allows us to bypass dealing with complex calculations involving large numbers by focusing directly on the exponents.

Understanding how to apply exponent rules is critical in solving exponential equations efficiently. Other important rules, though not directly used in this exercise, include the product of powers rule, power of a power rule, and quotient of powers rule. These rules together provide a comprehensive toolkit for simplifying and solving equations involving exponents.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying equations through operations such as addition, subtraction, multiplication, division, and factoring. In our example, after we equated the exponents, we received a linear equation: \(7 - x = 2x + 1\).

To solve this equation, we need to isolate \(x\). Here’s how you do it:
  • Add \(x\) to both sides: \(7 = 3x + 1\).
  • Subtract 1 from both sides to get: \(6 = 3x\).
  • Finally, divide by 3, resulting in: \(x = 2\).


Through these manipulations, we move terms across the equation, combine like terms, and apply operations strategically to solve for the unknown variable. Mastering these techniques allows us to handle not only simple linear equations but much more complex algebraic expressions as well.

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Most popular questions from this chapter

In 1840 , Britain experienced a bovine (cattle and oxen) epidemic called epizooty. The estimated number of new cases every 28 days is listed in the table. At the time, the London Daily made a dire prediction that the number of new cases would continue to increase indefinitely. William Farr correctly predicted when the number of new cases would peak. Of the two functions $$\begin{array}{l}f(t)=653(1.028)^{t} \\\g(t)=54,700 e^{-(t-200)^{2} / 7500}\end{array}$$ and one models the newspaper's prediction and the other models Farr's prediction, where \(t\) is in days with \(t=0\) corresponding to August \(12,1840\). $$\begin{array}{|c|c|}\hline \text { Date } & \text { New cases } \\\\\hline \text { Aug. 12 } & 506 \\\\\hline \text { Sept. 9 } & 1289 \\\\\hline \text { Oct. 7 } & 3487 \\\\\hline \text { Nov. 4 } & 9597 \\\\\hline \text { Dec. 2 } & 18,817 \\\\\hline \text { Dec. 30 } & 33,835 \\\\\hline \text { Jan. 27 } & 47,191 \\\\\hline\end{array}$$ (a) Graph each function, together with the data, in the viewing rectangle \([0,400,100]\) by \([0,60,000,10,000]\) (b) Determine which function better models Farr's prediction. (c) Determine the date on which the number of new cases peaked.

Solve the equation graphically. $$x \log x-\log x=5$$

(a) Prove that the function defined by \(f(x)=a x+b(a\) linear function) for \(a \neq 0\) has an inverse function, and find \(f^{-1}(x)\) (b) Does a constant function have an inverse? Explain.

Graph \(f\) on the given interval. (a) Determine whether \(f\) is one-to-one. (b) Estimate the zeros of \(f\). $$f(x)=\frac{3.1^{x}-2.5^{-x}}{2.7^{x}+4.5^{-x}} ; \quad[-3,3]$$

If monthly payments \(p\) are deposited in a savings account paying an annual interest rate \(r,\) then the amount \(A\) in the account after \(n\) years is given by $$A=\frac{p\left(1+\frac{r}{12}\right)\left[\left(1+\frac{r}{12}\right)^{12 n}-1\right]}{\frac{r}{12}}$$ Graph \(A\) for each value of \(p\) and \(r,\) and estimate \(n\) for \(A=100,000 \text{dollars}\). $$p=100, \quad r=0.05$$

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