# How do you solve e^(-x) = 5^(2x) ?

Dec 13, 2015

Apply properties of logarithms to find that $x = 0$

#### Explanation:

We will use the following properties:

• $\ln \left({a}^{x}\right) = x \ln \left(a\right)$

• $\ln \left(e\right) = 1$

${e}^{- x} = {5}^{2 x}$

$\implies \ln \left({e}^{- x}\right) = \ln \left({5}^{2 x}\right)$

$\implies - x \ln \left(e\right) = 2 x \ln \left(5\right)$

$\implies - x = 2 \ln \left(5\right) x$

$\implies 2 \ln \left(5\right) x + x = 0$

$\implies x \left(2 \ln \left(5\right) + 1\right) = 0$

$\implies x = 0$
The text passage contains a mathematical question and provides a detailed step-by-step solution to the problem.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# How do you solve e^(-x) = 5^(2x) ?

Dec 13, 2015

Apply properties of logarithms to find that $x = 0$

#### Explanation:

We will use the following properties:

• $\ln \left({a}^{x}\right) = x \ln \left(a\right)$

• $\ln \left(e\right) = 1$

${e}^{- x} = {5}^{2 x}$

$\implies \ln \left({e}^{- x}\right) = \ln \left({5}^{2 x}\right)$

$\implies - x \ln \left(e\right) = 2 x \ln \left(5\right)$

$\implies - x = 2 \ln \left(5\right) x$

$\implies 2 \ln \left(5\right) x + x = 0$

$\implies x \left(2 \ln \left(5\right) + 1\right) = 0$

$\implies x = 0$
user
The text passage contains a mathematical question and provides a detailed step-by-step solution to the problem.

TYPE: SINGLE_QUESTION_WITH_ANSWER