Problem involving change of base of Logarithm

[chijioke]

Solve the equation
$$9 \log_{ x }{ 5 } = \log_{ 5 }{ x }$$

Solution​

$$\log_{ x } { 5 }^{9} = \log_{ 5 }{ x }$$Changing the base of the left term,
we have
$$\log_{ x }{ 5 ^ { 9 }} = \frac{ \log_{ 5 }{ 5 ^ { 9 }} }{ \log_{ 5 }{ x } }$$Equating it to the right, we have
$$\frac{ \log_{ 5 }{ 5 ^ { 9 }} }{ \log_{ 5 }{ x } }= \log_{ 5 }{ x }$$ $$\frac{ \log_{ 5 }{ 5 ^ { 9 }} }{ \log_{ 5 }{ x } }= \frac{ \log_{ 5 }{ x } }{ 1 }$$
Cross multiplying, we have
$$\log_{ 5 }{ 5 ^ { 9 }} = \log_{ 5 }{ x } \times \log_{ 5 }{ x }$$I am stucked here. What am I doing right?
If I choose to go the other way, that is changing the base of the right term,
$$9 \log_{ x }{ 5 } = \log_{ 5 }{ x }$$$$\log_{ 5 }{ x } = \frac{ \log_{ x }{ x } }{ \log_{ x }{ 5 } }$$$$\frac{ 9 \log_{ x }{ 5 } }{ 1 } = \frac{ \log_{ x }{ x } }{ \log_{ x }{ 5 } }$$$$\frac{ 9 \log_{ x }{ 5 } }{ 1 } = \frac{ 1 }{ \log_{ x }{ 5 } }$$$$1 \times 1 = \log_{ x }{ 5 ^ { 9 }} \times \log_{ x }{ 5 }$$ $$\log_{ x }{ 5 ^ { 9 }} \left( \log_{ x }{ 5 } \right) = 1$$I am still stucked. I think need help.

 

[Steven G]

From your last line let u = log_x5 to get 9u²=1

I also suggest that you never cross multiply!

 

[Dr.Peterson]

Solve the equation
$$9 \log_{ x }{ 5 } = \log_{ 5 }{ x }$$

Solution​

$$\log_{ x } { 5 }^{9} = \log_{ 5 }{ x }$$Changing the base of the left term,
we have
$$\log_{ x }{ 5 ^ { 9 }} = \frac{ \log_{ 5 }{ 5 ^ { 9 }} }{ \log_{ 5 }{ x } }$$Equating it to the right, we have
$$\frac{ \log_{ 5 }{ 5 ^ { 9 }} }{ \log_{ 5 }{ x } }= \log_{ 5 }{ x }$$ $$\frac{ \log_{ 5 }{ 5 ^ { 9 }} }{ \log_{ 5 }{ x } }= \frac{ \log_{ 5 }{ x } }{ 1 }$$
Cross multiplying, we have
$$\log_{ 5 }{ 5 ^ { 9 }} = \log_{ 5 }{ x } \times \log_{ 5 }{ x }$$I am stucked here. What am I doing right?

Using that first way, what is the value of $\log_{5}{5^{9}}$? Then continue.

 

[blamocur]

From your last line let u = logx5 to get 9u2=1

... or even from the first line.

 

[pka]

Solve the equation
$$9 \log_{ x }{ 5 } = \log_{ 5 }{ x }$$

Solution​

$$\log_{ x } { 5 }^{9} = \log_{ 5 }{ x }$$ Changing the base of the left term

The problem is equivalent to: $\dfrac{9\log(5)}{\log(x)}=\dfrac{\log(x)}{\log(5)}$
Continue

 

[JeffM]

I must be missing something. You want to get the logs to the same base. Great, good thinking. Do it first. But once you have $\log_5(5)$, that reduces to 1. Why mess around when that obvious simplification is staring you in the face?

$$9 \log_x (5) = \log_5(x) \implies 9 * \dfrac{\log_5(5)}{\log_5 (x)} = log_5(x) \implies \dfrac{9}{\log_5(x)} = \log_5 (x) \implies \\ 9 = \{\log_5(x)\}^2 > 0.$$
Now you could do a u-substitution, but you could directly take the square roots of both sides of the equation. Then what?

 

[chijioke]

Using that first way, what is the value of $\log_{5}{5^{9}}$? Then continue.

$$\frac{ \log_{ 5 }{ 5 ^ { 9 }} }{ \log_{ 5 }{ x } } = \frac{ \log_{ 5 }{ x } }{ 1 }$$$$\frac{ 9 }{ \log_{ 5 }{ x } } = \frac{ \log_{ 5 }{ x } }{ 1 }$$Where do I go from here?

 

[JeffM]

$$\frac{ \log_{ 5 }{ 5 ^ { 9 }} }{ \log_{ 5 }{ x } } = \frac{ \log_{ 5 }{ x } }{ 1 }$$$$\frac{ 9 }{ \log_{ 5 }{ x } } = \frac{ \log_{ 5 }{ x } }{ 1 }$$Where do I go from here?

Did you read the peceding pst?

 

[chijioke]

Did you read the peceding pst?

Okay, here I go,
Let $$u= \log_{ 5 }{ x }$$$$\frac{ 9 }{ \log_{ 5 }{ x } } = \frac{ \log_{ 5 }{ x } }{ 1 }$$$$\frac{ 9 }{ u }= \frac{ u}{ 1 }$$Rearranging
$$u^ {2 } = 9$$$$u =3$$But $$\log_{ 5 }{ x } = 3$$$$x=5^3$$$$\therefore ~x= 125$$Applying the same thing to the right term,
let$$m = \log_{ x }{ 5 }$$$$1×1 = \log_{ x }{ 5 ^ { 9 }} \times \log_{ x }{ 5 }$$$$9m^2 =1$$$$m = \frac{ 1 }{ 3 }$$Putting m back for $$\log_{ x }{ 5 }\text{,~we have}$$$$\log_{ x }{ 5 } = \frac{ 1 }{ 3 }$$$$x ^ {\frac{ 1 }{ 3 }} = 5$$Thus, $$x = 125$$I think should be correct now. But do we have two values for x? I am asking this because I used my calculator to check and I got
$$\begin{matrix} x = 0.008 \\ x = 125 \end{matrix}$$If so, how is that supposed to be?

 

[Steven G]

If u^2=9, then u= +/- 3, not just 3!

 

[chijioke]

If u^2=9, then u= +/- 3, not just 3!

Yes that is true.

 

[pka]

Solve the equation $$9 \log_{ x }{ 5 } = \log_{ 5 }{ x }$$

To chijoke. you are hard-headed are you not?
$9\log_x(5)=\log_5(x)\\9\dfrac{\log(5)}{\log(x)}=\dfrac{\log(x)}{\log(5)}\\9\left[\log(5)\right]^2=\left[\log(x)\right]^2\\ 3\log(5)=\log(x)\text{using the square root}\\x=125$

SEE HERE

 

[Steven G]

If u^2=9, then u= +/- 3, not just 3!

Yes that is true.

So is there a solution for u=-3? Why?

 

[JeffM]

It is perhaps useful to elaborate the point indicated by Steven because bases below 1 for logarithms may be tricky for beginners. I suspect that pka is correct that only the positive answer is expected, but Steven is perfectly correct that there is a valid solution with the negative root.

$$9 = \{\log_5(x)\}^2 \implies log_5(x) = \pm 3 \implies x = 5^3 = 125 \text { or } x = 5^{-3} = \dfrac{1}{5^3} = \dfrac{1}{125} = 0.008. \\ \text {Checking answer from positive root: }\\ x = 125 \implies 9 \log_x(5) = 9 \log_{125} ( 125^{1/3}) = 9 * \dfrac{1}{3} * \log{125}(125) = 3 * 1 = 3.\\ \log_5(x) = \log_5(125) = \log_5(5^3) = 3 \log_5(5) = 3 * 1 = 3. \ \checkmark$$
Let’s explore when the base is between zero and one.

$$0 < a < 1 \text { and } 0 < b \implies \dfrac{1}{a} > 0 \implies \log_a(1/a) = - 1.\\ \therefore \log_{1/a}(b) = \dfrac{\log_a(b)}{\log_a(1/a)} =\dfrac{\log_a(b)}{-1} = - \log_a(b).$$
The log of any number to base 1/a (provided a is positive and not 1) is the additive inverse of the log of the same number to a. Now we can address the other solution.

$$x = 0.008 = (1/125) \implies 9 \log_{1/125}{5} = 9 \{- \log_{125}(5)\} = -9 * \log_{125}(5) = -9 * \dfrac{1}{3} = - 3.\\ \log_5 \left ( \dfrac{1}{125} \right ) = \log_5(125^{-1}) = - \log_5(125) = -3. \ \checkmark.$$

 

[Steven G]

It is perhaps useful to elaborate the point indicated by Steven because bases below 1 for logarithms may be tricky for beginners. I suspect that pka is correct that only the positive answer is expected, but Steven is perfectly correct that there is a valid solution with the negative root.

$$9 = \{\log_5(x)\}^2 \implies log_5(x) = \pm 3 \implies x = 5^3 = 125 \text { or } x = 5^{-3} = \dfrac{1}{5^3} = \dfrac{1}{125} = 0.008. \\ \text {Checking answer from positive root: }\\ x = 125 \implies 9 \log_x(5) = 9 \log_{125} ( 125^{1/3}) = 9 * \dfrac{1}{3} * \log{125}(125) = 3 * 1 = 3.\\ \log_5(x) = \log_5(125) = \log_5(5^3) = 3 \log_5(5) = 3 * 1 = 3. \ \checkmark$$
Let’s explore when the base is between zero and one.

$$0 < a < 1 \text { and } 0 < b \implies \dfrac{1}{a} > 0 \implies \log_a(1/a) = - 1.\\ \therefore \log_{1/a}(b) = \dfrac{\log_a(b)}{\log_a(1/a)} =\dfrac{\log_a(b)}{-1} = - \log_a(b).$$
The log of any number to base 1/a (provided a is positive and not 1) is the additive inverse of the log of the same number to a. Now we can address the other solution.

$$x = 0.008 = (1/125) \implies 9 \log_{1/125}{5} = 9 \{- \log_{125}(5)\} = -9 * \log_{125}(5) = -9 * \dfrac{1}{3} = - 3.\\ \log_5 \left ( \dfrac{1}{125} \right ) = \log_5(125^{-1}) = - \log_5(125) = -3. \ \checkmark.$$

JeffM,
You need to be careful as part of the original problem had log_x5, so x = -3 is not a possibility.

 

[JeffM]

JeffM,
You need to be careful as part of the original problem had log_x5, so x = -3 is not a possibility.

You are correct (as was pka). I am off to spend three hours in the corner.

 

[JeffM]

Wait a minute. Part of the original problem was that x was a base, and a base cannot be negative. True. I allowed my respect for Steven and pka to override my confidence in my own logic. I did not say that x was negative. I said that

$$9 = \{\log_5(x)\}^2 \implies\\ \log_5(x) = \pm 3 \implies x = 5^3 = 125 > 0 \text { or } x = 5^{-3} = 0.008 > 0.$$
There is no bar to the exponent of a positive number being negative. And 0.008 is indubitably a positive number.

In Steven’s notation $u = \pm 3$, and u is a logarithm. There is no bar to a logarithm being negative even though its argument must be positive.

Upon rereading pka’s answer, I see that he carefully noted that x = 125 when using the square root function, which is defined to be non-negative. He simply did not opine on the alternative root.

As for lookagain’s post its apparent logic is beyond silly. He starts from $\pm 3 \log(5) = \log(x).$ He did not show his work, but he then apparently proceeded

$$\pm 3 \log(5) = \log(x) \text { and } x > 0 \implies \\ \{x > 0 \text { and } \{3 \log(5) = \log(x)\} \text { or } \{- 3 \log(5) = \log(x) \text { and } x > 0\} \implies\\ \{x > 0 \text { and } \{\log(5^3) = \log(x)\} \text { or } \{ \log(5^{-3} = \log(x) \text { and } x > 0\} \implies \\ \{x = 5^3 \text { and } x > 0\} \text { or } \{x = 5^{-3} \text { and } x > 0\} \implies \\ \{x = 125 \text { and } x > 0\} \text { or } \{x = 0.008 \text { and } x > 0\}.\\ \text {But } 0.008 \le 0.\\ \therefore \ x = 125, \text {which is the unique solution.}$$
There are two valid solutions, just as the OP’s calculator said.

 

[lookagain]

I reported my post # 15 for being unhelpful.

 

[lookagain]

As for lookagain’s post its apparent logic is beyond silly.

Reply to me directly in the thread, and as soon as possible, instead of talking around me. Also, please keep figurative words
such as "silly" to a limit, because it was easier for you to write that when you talked around me, instead of directly to me.
Then, I will have a greater chance of correcting my post properly.

 

[Steven G]

Wait a minute. Part of the original problem was that x was a base, and a base cannot be negative. True. I allowed my respect for Steven and pka to override my confidence in my own logic. I did not say that x was negative. I said that

$$9 = \{\log_5(x)\}^2 \implies\\ \log_5(x) = \pm 3 \implies x = 5^3 = 125 > 0 \text { or } x = 5^{-3} = 0.008 > 0.$$
There is no bar to the exponent of a positive number being negative. And 0.008 is indubitably a positive number.

In Steven’s notation $u = \pm 3$, and u is a logarithm. There is no bar to a logarithm being negative even though its argument must be positive.

Upon rereading pka’s answer, I see that he carefully noted that x = 125 when using the square root function, which is defined to be non-negative. He simply did not opine on the alternative root.

As for lookagain’s post its apparent logic is beyond silly. He starts from $\pm 3 \log(5) = \log(x).$ He did not show his work, but he then apparently proceeded

$$\pm 3 \log(5) = \log(x) \text { and } x > 0 \implies \\ \{x > 0 \text { and } \{3 \log(5) = \log(x)\} \text { or } \{- 3 \log(5) = \log(x) \text { and } x > 0\} \implies\\ \{x > 0 \text { and } \{\log(5^3) = \log(x)\} \text { or } \{ \log(5^{-3} = \log(x) \text { and } x > 0\} \implies \\ \{x = 5^3 \text { and } x > 0\} \text { or } \{x = 5^{-3} \text { and } x > 0\} \implies \\ \{x = 125 \text { and } x > 0\} \text { or } \{x = 0.008 \text { and } x > 0\}.\\ \text {But } 0.008 \le 0.\\ \therefore \ x = 125, \text {which is the unique solution.}$$
There are two valid solutions, just as the OP’s calculator said.

The next time you need to go to the corner, I'll go there for you.