/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Solve the equation. $$27^{x-1}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the equation. $$27^{x-1}=9^{2 x-3}$$

Short Answer

Expert verified
The solution is \(x = 3\).

Step by step solution

01

Express in terms of the same base

To solve the equation \(27^{x-1} = 9^{2x-3}\), we first need to express both sides using the same base. Notice that 27 and 9 can both be expressed as powers of 3:- \(27 = 3^3\), so \(27^{x-1} = (3^3)^{x-1} = 3^{3(x-1)}\).- \(9 = 3^2\), so \(9^{2x-3} = (3^2)^{2x-3} = 3^{2(2x-3)}\).
02

Simplify the exponents

Simplifying the expressions by using the power of a power rule, which states that \((a^m)^n = a^{mn}\):- Expression for \(27^{x-1}\): \[3^{3(x-1)} = 3^{3x-3}\].- Expression for \(9^{2x-3}\): \[3^{2(2x-3)} = 3^{4x-6}\].
03

Set the exponents equal

Since the bases are the same, we can set the exponents equal to each other to solve for \(x\):\[3x - 3 = 4x - 6\].This follows from the property that if \(a^m = a^n\), then \(m = n\) for any \(a eq 0\).
04

Solve for x

Solve the equation \(3x - 3 = 4x - 6\):1. Subtract \(3x\) from both sides: \[-3 = x - 6\].2. Add 6 to both sides: \[3 = x\].
05

Verify the solution

Substitute \(x = 3\) back into the original expression to verify:1. Calculate \(27^{3-1} = 27^2 = 729\). 2. Calculate \(9^{2(3)-3} = 9^3 = 729\).Since both sides equal 729, the solution \(x = 3\) is verified as correct.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Expressing Expressions with Common Base
When we're faced with exponential equations involving different bases, the first step is often to rewrite each expression using a common base. This helps streamline the equation, making it easier to solve. Let's take an example with the numbers 27 and 9. Both of these numbers can be expressed as powers of 3.
  • 27 can be rewritten as \(3^3\)
  • 9 can be written as \(3^2\)
By expressing each side of the equation \(27^{x-1} = 9^{2x-3}\) with a base of 3, we transform it into \((3^3)^{x-1} = (3^2)^{2x-3}\).
This transformation is crucial because it allows us to focus on the exponents, simplifying our work when trying to find a solution.
Power of a Power Rule
The power of a power rule is a fundamental principle in solving exponential equations that involve expressions such as \((a^m)^n\). This rule states that you multiply the exponents: \((a^m)^n = a^{mn}\).
This means if you have \((3^3)^{x-1}\), you can simplify it to \(3^{3(x-1)}\) or \(3^{3x - 3}\). Similarly, \((3^2)^{2x-3}\) simplifies to \(3^{2(2x-3)}\) or \(3^{4x - 6}\).
This step is about simplifying the exponential expressions so all the complexities of the problem are boiled down to a straightforward equation involving multiplied exponents. This sets the scene for easily equating the two sides and solving for the unknown variable.
Solving for Variables
Once you have rewritten the exponential equation with a common base and simplified their exponents using the power of a power rule, the next task is to solve for the variable. Since the bases are the same, you can set the exponents equal to each other. For example:
  • With bases of 3 consolidated, you have \(3^{3x - 3} = 3^{4x - 6}\)
  • By the property of exponents, you equate: \(3x - 3 = 4x - 6\)
You solve this linear equation by rearranging terms:
  • Subtract \(3x\) from both sides to get \(-3 = x - 6\)
  • Add 6 to both sides to isolate \(x\), resulting in \(3 = x\)
Thus, the value of \(x\) is found to be 3. This process involves basic algebraic manipulation and is much more straightforward once the expressions are simplified.
Verification of Solution
No solution is complete without verification. To ensure the result is accurate, substitute the found value of the variable back into the original exponential equation.
For this problem, our solution is \(x = 3\). Let's check:
  • First, calculate \(27^{3-1} = 27^2\). Since 27 squared is 729, the left side becomes 729.
  • Next, calculate \(9^{2(3)-3} = 9^3\). Nine cubed is also 729, so the right side becomes 729 too.
With both sides equating to 729, our original equation is verified. This step is crucial for ensuring that any possible mistakes during the simplification or solving process are caught, confirming \(x = 3\) as the true solution.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Probability density function In statistics, the probability density function for the normal distribution is defined by $$ f(x)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-z^{2 / 2}} \quad \text { with } \quad z=\frac{x-\mu}{\sigma} $$ where \(\mu\) and \(\sigma\) are real numbers ( \(\mu\) is the mean and \(\sigma^{2}\) is the variance of the distribution). Sketch the graph of \(f\) for the case \(\sigma=1\) and \(\mu=0\).

If \(v\) denotes the wind velocity (in \(\mathrm{m} / \mathrm{sec}\) ) at a height of \(z\) meters above the ground, then under certain conditions \(v=c \ln \left(z / z_{0}\right)\), where \(c\) is a positive constant and \(z_{0}\) is the height at which the velocity is zero. Sketch the graph of this equation on a \(z v\)-plane for \(c=0.5\) and \(z_{0}=0.1 \mathrm{~m}\).

If \(p\) denotes the selling price (in dollars) of a commodity and \(x\) is the corresponding demand (in number sold per day), then the relationship between \(p\) and \(x\) is sometimes given by \(p=p_{0} e^{-a x}\), where \(p_{0}\) and \(a\) are positive constants. Express \(x\) as a function of \(p\).

Solve the equation. $$ 2 \log _{3} x=3 \log _{3} 5 $$

A drug is eliminated from the body through urine. Suppose that for a dose of 10 milligrams, the amount \(A(t)\) remaining in the body \(t\) hours later is given by \(A(t)=10(0.8)^{t}\) and that in order for the drug to be effective, at least 2 milligrams must be in the body. (a) Determine when 2 milligrams is left in the body. (b) What is the half-life of the drug?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.