Logarithms question

[jayson]

log(4x) + log(x) = 2.459

is this done properly the following way?


antilog (4x) + antilog (x) = antilog 2.459

4x + x = 287.7398 ?

 

[mmm4444bot]

Antilogs? Are you using a slide rule?

I would start by combining the two lograrithmic terms.

There is a property of logarithms for this.

log(A) + log(B) = log(A*B)

Apply that, switch to exponential form, and you'll have an easy quadratic equation to solve for x.

Easy using a calculator, that is. :cool:

 

[chowe13]

wow, antilog huh, you are making it more difficult than it needs to be,

Change to log (4x(x)) = 2.49
then you have log 4x²=2.49
eliminate log by raising both sides to the power of 10, this gives you

4x²=10^2.49

Then just solve for x, divide by 4, take the square root, and bam, done.

 

[jayson]

I think I didn't have enough sleep when I started these exercises but I am good to go. I finished all of my exercises and I got all of them right except this one.



3(10^s)=11-2(10^s)



Is there anything you guys can do to explain how to solve for this one?

 

[lookagain]

\(\displaystyle \text{2.459 is actually the correct number as given by the OP:}\)

Change to log (4x(x)) = 2.459


then you have log 4x² = 2.459


eliminate log by raising both sides to the power of 10, this gives you


4x² = 10^2.459


Then just solve for x, divide by 4, take the square root, and bam, done.


\(\displaystyle \text{No. } \) \(\displaystyle . \ \ \ \ \ \ \ x^2 = \dfrac{10^{2.459}}{4}}\)


\(\displaystyle x = \pm \sqrt{\dfrac{10^{2.459}}{4}}\)


\(\displaystyle \text{A solver gets two candidate solutions from this and must }\)

\(\displaystyle \text{check them into the original equation to see which ones }\)

\(\displaystyle \text{(if any) work.}\)

.

 

[mmm4444bot]

\(\displaystyle \text{A solver gets two candidate solutions from this and must check them}\)

I wish that I had remembered to say this. :cool:

Checking the answer is very important, when a solver has taken the square root of both sides of an equation.

 

[jayson]

Thanks guys! Thanks for making things obvious tot this math bum.