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.