Find x

[frctl]

I don't know how to solve this using the properties of logarithmic functions.

 

[Steven G]

Can you write each log term on the right side of the equal sign without a multiple in front.

For example 3log₄(9) = log₄(9³)

Can you then write it with one log?

 

[JeffM]

What jomo is suggesting is quicker, but you could also think about clearing fractions.

[MATH]log_b(x) = \frac{2}{3} log_b(8) + \frac{1}{2} log_b(9) - log_b(6) \implies [/MATH]
[MATH]2 * 3 * log_b(x) = 2 * 3 * \left (\frac{2}{3} log_b(8) + \frac{1}{2} log_b(9) - log_b(6) \right ) \implies[/MATH]
[MATH]6log_b(x) = 4 log_b(8) + 3 log_b(9) - 6log_b(6) \implies WHAT?[/MATH]

 

[frctl]

Without a multiple in front

Attempt to write as one log

 

[JeffM]

Without a multiple in front
View attachment 16997

Perfect. Going this way, I'd suggest simplifying the arguments of the logs next, but it is not necessary.

But then you lost your way BIG TIME

Attempt to write as one log
View attachment 16998

[MATH]log_b(y) + log_b(z) \ne log_b(y + z).[/MATH][MATH] \text { INSTEAD, } log_b(y) + log_b(z) = \text {WHAT?}[/MATH]

 

[frctl]

How do I write

As one log

 

[pka]

What jomo is suggesting is quicker, but you could also think about clearing fractions.
[MATH]log_b(x) = \frac{2}{3} log_b(8) + \frac{1}{2} log_b(9) - log_b(6) \implies [/MATH][MATH]2 * 3 * log_b(x) = 2 * 3 * \left (\frac{2}{3} log_b(8) + \frac{1}{2} log_b(9) - log_b(6) \right ) \implies[/MATH][MATH]6log_b(x) = 4 log_b(8) + 3 log_b(9) - 6log_b(6) \implies WHAT?[/MATH]

Hi Jeff you miss-groups..
\(\dfrac{2}{3}\log_b(8)=2\log_b(2)\) , \(\dfrac{1}{2}\log_b(9)=\log_b(3)\) and \(\log_b(6)=\log_b(2)+\log_b(3)\)

 

[JeffM]

Hi Jeff you miss-groups..
\(\dfrac{2}{3}\log_b(8)=2\log_b(2)\) , \(\dfrac{1}{2}\log_b(9)=\log_b(3)\) and \(\log_b(6)=\log_b(2)+\log_b(3)\)

Yes, that is easiest of all. Duh.

 

[JeffM]

How do I write
View attachment 16999
As one log

When you add the logs of numbers you get the log of what?

 

[frctl]

I obtained the following (not sure about the exponents)
- logb6

 

[JeffM]

[MATH]log_b(8^{2/3}) + log_b(9^{1/2} = log_b(8^{2/3} * 9^{1/2}) = log_b(4 * 3) = log_b(12).[/MATH]
Can you explain why? (HINT: Remember PEMDAS.)

Now [MATH]log_b(u) - log_b(v) = WHAT?[/MATH]

 

[frctl]

I'm not sure how you obtained logb(4 · 3)

Does it equal
logb(u · v)

 

[pka]

Without a multiple in front
View attachment 16997
Attempt to write as one log
View attachment 16998

To frctl, Do you know the properties of logarithms? Make it easy on yourself.
\( \dfrac{2}{3} \log_b(8) + \dfrac{1}{2} \log_b(9) - \log_b(6)=2\log_b(2)+\log_b(3)-\log_b(2)-\log_b(3)=\log_b(2)\)

 

[frctl]

The property I am looking for is
logbMN = logbM + logbN

I don't understand the last bit you wrote

 

[pka]

I don't understand the last bit you wrote

\(\dfrac{2}{3}\log_b(8)=2\log_b(2)\) If you do not follow that then please hire a face-to-face tutor.

 

[Deleted member 4993]

View attachment 17002

I'm not sure how you obtained logb(4 · 3)

Does it equal
logb(u · v)

NO!

log_b(u) - log_b(v) = log_b(u / v)

 

[Steven G]

I will work this out for you using pka's ideas.

\(\displaystyle \frac{2}{3}\)log_b8 = 2log_b8^1/3 = 2log_b2

log_b9^1/2 = log_b3

log_b6 = log_b(2*3) = log_b2 + log_b3

Actually you can continue from here.

 

[frctl]

How is this



Or I might have mixed up the signs

 

[JeffM]

I shall do this step by step, using pka's idea, slightly modified.

But first here are the basic laws of logarithms.

Given that a, b, and c are positive real numbers, and b is not equal to 1:

[MATH]log_b(a) = x \iff a = b^x[/MATH].

[MATH]\therefore log_b(1) = 0 \text { and } log_b(b) = 1.[/MATH]
[MATH]log_b(a * c) = log_b(a) + log_b(c).[/MATH]
[MATH]-\ log_b(a) = log_b \left ( \dfrac{1}{a} \right ) \implies log_b(c) - log_b(a) = log_b \left ( \dfrac{c}{a} \right ).[/MATH]
[MATH]u log_b(a) = log_b(a^u).[/MATH]
[MATH]log_b(a) = log_b(c) \iff a = c.[/MATH]
These along with the change of base law must be memorized and understood to use logs.

Now your problem step by step the easy way

[MATH]log_b(x) = \frac{2}{3} log_b(8) + \frac{1}{2} log_b(9) - log(6)[/MATH].

[MATH]\therefore log_b(x) = \frac{2}{3} log_b(2^3) + \frac{1}{2} log_b(3^2) - log_b(6)[/MATH].

[MATH]\therefore log_b(x) =log_b(2^{(3 * 2 /3)})+ log_b(3^{(2 * 1/2)}) - log_b(6).[/MATH]
[MATH]\therefore log_b(x) = log_b(2^2) + log_b(3^1) - log_b(6).[/MATH]
[MATH]\therefore log_b(x) = log_b(4) + log_b(3) - log_b(6).[/MATH]
[MATH]\therefore log_b(x) = log_b(4 * 3) - log_b(6) = log_b(12) - log_b(6).[/MATH]
[MATH]\therefore log_b(x) = log_b \left ( \dfrac{12}{6} \right ).[/MATH]
[MATH]\therefore log_b(x) = log_b(2).[/MATH]
[MATH]\therefore x = 2.[/MATH]
Learn the laws and be careful, and logs will give you no trouble.

 

[frctl]

Thank you