Log exponential and hyperbolics

[Sunny1982]

Hi Folks I was wondering if someone can be kind enough to see if the following answers are correct, I am working on rest of the questions of this assignment will be grateful if someone let's me know I'm on the right lines.
1. Solve the following:

3^x = 10
x=10/3

Ln (x²) + ln(x⁵)= 28
ln(x⁷) = 28
x⁷ = e^(28)
x = ⁷√(e^(28)
x = e^(4)

 

[JeffM]

Hi Folks I was wondering if someone can be kind enough to see if the following answers are correct, I am working on rest of the questions of this assignment will be grateful if someone let's me know I'm on the right lines.
1. Solve the following:

3^x = 10
x=10/3 NO! Let's consider this example \(\displaystyle 3^5 = 243, and \dfrac{243}{3} = 81 \ne 5.\)

Ln (x²) + ln(x⁵)= 28
ln(x⁷) = 28
x⁷ = e^(28)
x = ⁷√(e^(28)
x = e^(4) YES!

You can test your own work. \(\displaystyle ln\left\{(e^4)^2\right\} + \ln\left\{(e^4)^5\right\} = ln(e^8) + ln(e^{20}) = ln(e^{28}) = 28 * ln(e) = 28 * 1 = 28.\)

\(\displaystyle 3^x = 10 \implies ln(3^x) = ln(10) \implies x * ln(3) = ln(10) \implies x = \dfrac{ln(10)}{ln(3)}\)

 

[Sunny1982]

So in what format would I write the answer for this? 3^x = 10
x=10/3

 

[JeffM]

So in what format would I write the answer for this? 3^x = 10
x=10/3

The answer is NOT 10/3. So why would you want to write it in any format?

 

[Sunny1982]

Sorry to be rude so how do I work out the answer, that example didn't make sense to me?

 

[DrPhil]

\(\displaystyle 3^x = 10 \implies ln(3^x) = ln(10) \implies x * ln(3) = ln(10) \implies x = \dfrac{ln(10)}{ln(3)}\)

This answer was given in the first response to your question.

To bring the unknown x down from being an exponent, take the log of both sides of the equation and then solve for x. Do you need explanation of the steps?

 

[JeffM]

Sorry to be rude so how do I work out the answer, that example didn't make sense to me?

\(\displaystyle a = b \iff log_c(a) = log_c(b)\ for\ any\ real\ c > 0\ and\ c\ne 1.\)

In words: if two numbers are equal to each other, then their logs (to the same base) are equal; and if two logs to the same base are equal to each other, then their arguments are equal.

You with me?

So

\(\displaystyle 3^x = 10 \implies\)

\(\displaystyle log_c\left(3^x\right) = log_c(10), where\ c > 1.\) Just applying the general rule explained above.

And \(\displaystyle log_c\left(3^x\right) = x * log_c(3).\) One of the basic laws of logarithms, right?

So \(\displaystyle x * log_c(3) = log_c(10).\) Basic rule of algebra: two numbers equal to a third number are equal to each other.

And \(\displaystyle log_c(3) > 0 \implies log_c(3) \ne 0.\) So I can divide both sides of the equation by \(\displaystyle log_c(3).\)

\(\displaystyle x = \dfrac{log_c(10)}{log_c(3)}.\)

Now let's check our answer.

\(\displaystyle y = 3^x\ and\ x = \{log_c(10) \div log_c(3)\} \implies log_c(y) = log_c\left(3^{\{log_c(10) \div log_c(3)\}}\right) = \dfrac{log_c(10)}{log_c(3)} * log_c(3) = log_c(10).\)

\(\displaystyle log_c(y) = log_c(10) \implies y = 10 \implies 3^x = 10.\) It checks.

 

[lookagain]

\(\displaystyle a = b \iff log_c(a) = log_c(b)\ for\ any\ real\ c > 0.\)**

In words: if two numbers are equal to each other, then their logs (to the same base) are equal; and if two logs to the same base are equal to each other, then their arguments are equal.

**\(\displaystyle \ \ And \ \ c \ \ne \ 1.\)

 

[JeffM]

**\(\displaystyle \ \ And \ \ c \ \ne \ 1.\)

Thanks for the correction. I have fixed it.