Logarithmic Equations: 1 + 2e^3x = 7, 3log(4x) - 2 = 10

TOBYMAN

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Dec 9, 2008
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Can someone help me out with these two problems including the steps to the answer?

1+2e^3x=7

and

3log(4x)-2=10
 


These are basic equations involving properties of logarithms and exponents. Are you able to show any work or expain your reasoning so that we know why you're stuck?

Solve the first equation the same way as the examples in your textbook.

Isolate the term e^(3x) on one side of the equation.

Take the natural logarithm of both sides of this equation.

Use the property that allows you to remove the exponent 3x and relocate it as a factor, or simply use the definition of the natural logarithm of e^exponent to get exponent.

Solve for x.

Check your result in the original equation.

Solve the second equation the same way as the examples in your textbook.

Isolate log(4x) on one side of the equation.

Use the definition of log-base 10 to rewrite the equation using a power of ten (i.e., rewrite as an exponentiation).

Solve for x.

Check your result in the original equation.

If you need more help, then show your work, and try to say something about why you're stuck.

 
Re: Logarithmic Equations

Hello, TOBYMAN!


\(\displaystyle 1+2e^{3x} \:=\:7\)

\(\displaystyle \text{We have: }\;2e^{3x} \:=\:6 \quad\Rightarrow\quad e^{3x} \:=\:3\)

\(\displaystyle \text{Take logs: }\;\ln\left(e^{3x}\right) \:=\:\ln(3) \quad\Rightarrow\quad 3x\ln(e) \:=\:\ln(3)\)

\(\displaystyle \text{Since }\ln(e) = 1\text{, we have: }\;3x \:=\:\ln(3) \quad\Rightarrow\quad x \;=\;\tfrac{1}{3}\ln(3)\)



\(\displaystyle 3\log(4x)-2\:=\:10\)

\(\displaystyle \text{We have: }\;3\log(4x) \:=\:12 \quad\Rightarrow\quad \log(4x) \:=\;4\)

\(\displaystyle \text{Then: }\;4x \:=\:10^4 \:=\:10,\!000 \quad\Rightarrow\quad x \:=\:2,\!500\)

 
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