Express as a single longarithm

[caligirl350]

Express as a single logarithm

logb 5x+7(logbx-logby)

I came up with

(logb5x)(logbx^7)/logby^7

Can anyone tell me if this is correct?

 

[mmm4444bot]

(logb5x)(logbx^7)/logby^7

Can anyone tell me if this is correct?

It's not correct. The instructions ask for a single logarithm. Your result contains three logarithms.

You're on the right track, however; the properties show us to multiply or divide the logarithms' inputs, not to multiply or divide the logarithms themselves.

I'm using the word "input" to mean the logarithmic function's argument. In log_b(5x), the input is 5x.

Sum of two logarithms with same base: log_b(5x) + log_b(x^7) = log_b(5x * x^7)

Difference of two logarithms with same base: log_b(5x^8) - log_b(y^7) = log_b(5x^8/y^7)

Those results are single logarithms.

 

[JuicyBurger]

Express as a single logarithm

logb 5x+7(logbx-logby)

So, to get you started with what you have...

\(\displaystyle log_b(5x)+7(log_b(x)-log_b(y))\)

This can be re-written (using one of mmm4444bot's above methods)

\(\displaystyle log_b(5x)+7(log_b(\frac{x}{y}))\)

Now use the rule that mmm4444bot posted for adding logs and you should find your answer.

Hope this helps!

Remember that if you have something that looks like this:

\(\displaystyle 5log_b(x)\)

That it can be re-written like this:

\(\displaystyle log_b(x^5)\)

 

[caligirl350]

So my correct answer would be logb(5x^8/y^7)?

 

[JeffWest]

Looks great. Great use of the log rules.