Logs: Solving logarithmic equation log_4 x - log_4 6 = 3

[Gadsilla]

It's confusing me since there are two logs. (log4x-log4). If it was for example "log3(9x+4)=5, x =" then I'd know what to do. How different us it? Just for info, it's from an old test.

 

[Dr.Peterson]

It's confusing me since there are two logs. (log4x-log4). If it was for example "log3(9x+4)=5, x =" then I'd know what to do. How different us it? Just for info, it's from an old test.

The equation is

log₄ x - log₄ 6 = 3
​

Do you know how to condense two logs into a single log?

 

[Gadsilla]

The equation is

log₄ x - log₄ 6 = 3
​

Do you know how to condense two logs into a single log?

x=log4x(log4*6)=3

?

I can't find out how to make it like you did with the "4". And to the power of.

 

[Dr.Peterson]

x=log4x(log4*6)=3

?

I don't know what that means. What is the "*" for? And why two equal signs?

Please type equations carefully; for subscripts, either use the subscript button on the editor (x₂) or use the notation log_4(x), preceding the subscript with an underscore and putting the argument in parentheses for clarity.

The rule you need to use is that log(a/b) = log(a) - log(b).

 

[Dr.Peterson]

Have you read the Read Before Posting notice? That includes a link to "Formatting Math as Text", which can help you express yourself clearly without needing anything fancy.

 

[Gadsilla]

I don't know what that means. What is the "*" for? And why two equal signs?

Please type equations carefully; for subscripts, either use the subscript button on the editor (x₂) or use the notation log_4(x), preceding the subscript with an underscore and putting the argument in parentheses for clarity.

The rule you need to use is that log(a/b) = log(a) - log(b).

͏

 

[Dr.Peterson]

No, the answer is not what you wrote, "x = log₄ x ^{(log₄ 6)}".

Did you read what I said about the rule to use? Do you have a reason for writing what you did? We may need to discuss some misunderstanding.

 

[pka]

x=log4x(log4*6)=3 ?
I can't find out how to make it like you did with the "4". And to the power of.

Have you studied the logarithmic properties?
For example, are you conformable with \(\displaystyle \log_4(x)-\log_4(6)=\log_4\left(\dfrac{x}{6}\right)~?\)

Or do you understand that if \(\displaystyle \log_b(x)=a\text{ then }x=b^a~?\)

 

[Gadsilla]

No, the answer is not what you wrote, "x = log₄ x ^{(log₄ 6)}".

Did you read what I said about the rule to use? Do you have a reason for writing what you did? We may need to discuss some misunderstanding.

?

 

[Dr.Peterson]

You wrote,

log_4(x/6) = 3

Now take the next step, which pka already mentioned to you.

 

[Gadsilla]

I have no idea where to go from there.

 

[Dr.Peterson]

Have you studied the logarithmic properties?
For example, are you conformable with \(\displaystyle \log_4(x)-\log_4(6)=\log_4\left(\dfrac{x}{6}\right)~?\)

Or do you understand that if \(\displaystyle \log_b(x)=a\text{ then }x=b^a~?\)

You wrote,

log_4(x/6) = 3

Now take the next step, which pka already mentioned to you.

I have no idea where to go from there.

Oh?

You were reminded of the fact that if log_b(x) = a, then x = b^a. (We use "^" to indicate an exponent when we don't have formatting available, or choose not to use it.) This is essentially the definition of the logarithm.

In your problem, "b" is 4, "x" is x/6, and "a" is 3. So what is the new equation you can write using this rule?

 

[Gadsilla]

Oh?

You were reminded of the fact that if log_b(x) = a, then x = b^a. (We use "^" to indicate an exponent when we don't have formatting available, or choose not to use it.) This is essentially the definition of the logarithm.

In your problem, "b" is 4, "x" is x/6, and "a" is 3. So what is the new equation you can write using this rule?

I think I get it.

 

[Dr.Peterson]

I think I get it.

One more step and you're finished.

 

[Gadsilla]

One more step and you're finished.

I would divide by 6, but I can't ? What ?

 

[Dr.Peterson]

I would divide by 6, but I can't ? What ?

Why do you want to divide?

You have x/6 = 64. How do you undo a division by 6?

 

[pka]

I would divide by 6, but I can't ? What ?

\(\displaystyle \begin{align*}\dfrac{x}{6}&=64 \\x&=64\cdot 6 \end{align*}\)

What is hard to understand about that? Please tell us in detail.

 

[Gadsilla]

\(\displaystyle \begin{align*}\dfrac{x}{6}&=64 \\x&=64\cdot 6 \end{align*}\)

What is hard to understand about that? Please tell us in detail.

But if it was 6/x = 64 you would divide ?

 

[pka]

But if it was 6/x = 64 you would divide ?

\(\displaystyle \begin{align*}\dfrac{6}{x}&=\dfrac{64}{1}\\\ \; \\\dfrac{x}{6}&=\dfrac{1}{64}\\ \; \\x&=\dfrac{6}{64} \end{align*}\)

 

[Steven G]

But if it was 6/x = 64 you would divide ?

You simply interchange the x and the 64 to get x = 6/64