Writing equation in a form not involving logarithm

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chijioke

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Write the equation
[math]2 \log_{ 3 }{ x } + x \log_{ 3 }{ (b + c) } - y ^ { 2 } = 3[/math]in a form not involving logarithm.
Solution​
[math]2 \log_{ 3 }{ x } + x \log_{ 3 }{ (b + c) } - y ^ { 2 } = 3 \\ \implies[/math][math]\log_{ 3 }{ x ^ { 2 }} + \log_{ 3 }{ (b + c )} ^ { x } - y ^ { 2 } = 3[/math]Am stucked. What do I do next?
 
Harry_the_cat said:
Add \(\displaystyle y^2\) to both sides, then combine logs. Can you see what to do next?
Click to expand...
That is
Solution​
[math]2 \log_{ 3 }{ x } + x \log_{ 3 }{ (b + c) } - y ^ { 2 } = 3 \\ \implies[/math]
[math]\log_{ 3 }{ x ^ { 2 }} + \log_{ 3 }{ (b + c )} ^ { x } - y ^ { 2 } = 3[/math]​
[math]\implies[/math] [math]\log_{ 3 }{ x ^ { 2 }} + \log_{ 3 }{ b + c } ^ { x } = 3 + y ^ { 2 }[/math][math]\implies[/math][math]\log_{ 3 }{ x ^ { 2 }} ( b + c )^ { x } = 3 + y ^ { 2 }[/math]Writing [math]3 ^ { (3 + y ^ { 2 } ) } = x ^ { 2 } (b + c )[/math]Would be wrong. Wouldn't it? So what do I do next?
 
No, it is not wrong. Why do you think it is?

EDIT: You skipped some steps, which may be why you doubt yourself.

[math] \log_3\{x^2(b + c)^x\} = 3 + y^2 = (3 + y^2) * 1 = (3 + y^2) * \log_3(3) = \log_3(3^{3 + y^2)} \implies \\ x^2(b + c)^x = 3^{3 + y^2)} [/math]
 
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chijioke said:
Write the equation
[math]2 \log_{ 3 }{ x } + x \log_{ 3 }{ (b + c) } - y ^ { 2 } = 3[/math]
Click to expand...
Given [imath]2\log_{3}(x)+x\log_{3}(a+b)-y^2=3\\\log_{3}(x^2)+\log_{3}(a+b)^x=y^2+3[/imath]
[imath] \log_{3}\left\{(x^2)(a+b)^x \right\}=y^2+3\\[/imath]
[imath]3^{y^2+3}=x^2(a+b)^x[/imath]

We have the same result.
All you ask for is that it be written without logs.
 
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JeffM said:
No, it is not wrong. Why do you think it is?
[math] \log_3\{x^2(b + c)^x\} = 3 + y^2 = (3 + y^2) * 1 = (3 + y^2) * \log_3(3) = \log_3(3^{3 + y^2)} \implies \\ x^2(b + c)^x = 3^{3 + y^2)} [/math]
Click to expand...
I think it is wrong because of the right side [math]3^{(3 + y^2)}[/math]Since I began solving mathematical problem, I have never seen index upon index. What I mean by that is the 3 is already raised to the power of [math](3 + y^2)[/math]. You can see that the two terms inside the bracket are power of 3 and right inside the bracket, the y is a base carrying power 2. That is power upon power. To me, is like a baby carrying a baby. I have never seen that kind of math before. Have you ever seen that kind of math problem before? If you have, could you please show me examples?
 
What if y=2? Then 3^(3+y^2) = 3^(3+2^2) = 3^(3+4) = 3^7= figure that out on your own

You can raise 3 to any power, including 3.
 
pka said:
Given [imath]2\log_{3}(x)+x\log_{3}(a+b)-y^2=3\\\log_{3}(x^2)+\log_{3}(a+b)^x=y^2+3[/imath]
[imath] \log_{3}\left\{(x^2)(a+b)^x \right\}=y^2+3\\[/imath]
[imath]3^{y^2+3}=x^2(a+b)^x[/imath]

We have the same result.
All you ask for is that it be written without logs.
Click to expand...
BTW HERE is a complete solution!
 
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chijioke said:
That is
Solution​
[math]2 \log_{ 3 }{ x } + x \log_{ 3 }{ (b + c) } - y ^ { 2 } = 3 \\ \implies[/math]
​
[math]\log_{ 3 }{ x ^ { 2 }} + \log_{ 3 }{ (b + c )} ^ { x } - y ^ { 2 } = 3[/math]​

[math]\implies[/math] [math]\log_{ 3 }{ x ^ { 2 }} + \log_{ 3 }{ b + c } ^ { x } = 3 + y ^ { 2 }[/math][math]\implies[/math][math]\log_{ 3 }{ x ^ { 2 }} ( b + c )^ { x } = 3 + y ^ { 2 }[/math]Writing [math]3 ^ { (3 + y ^ { 2 } ) } = x ^ { 2 } (b + c )[/math]Would be wrong. Wouldn't it? So what do I do next?
Click to expand...
Looks good except, you need brackets in your third line and you've lost a power of x in the last line. Other than that it is correct.
 
What do you mean by how do you make use of it?
You input values for a, b & x, and the output is the value for y.
If you mean how do you solve for y, then start with the following and post back with your work for further hints.

\(\displaystyle 3^{(y^2+3)} = 3^{y^2}3^3 = 27*3^{y^2}\)
Now divide both sides by 27 and ....
 
chijioke said:
I have been trying to make use of this your calculator but I have not succeeded for a long time now? How do you make use of it?
Click to expand...
Do you mean, how do you use Wolfram Alpha, the site @pka used?

If so, they provide examples of how to use it, such as this page.

If you mean something else, please be more specific. Also, you need to show us the actual problem you are trying to solve; is the goal to solve for y in terms of x, or something else? What you asked for explicitly does not lead to that goal.
 
chijioke said:
What did you type in that made the calculator to write the equation in a form not involving Logarithm?
Click to expand...
First of all that is an online program Wolfram Alpha SEE HERE

[imath]3^{y^2+3}=x^2(a+b)^x[/imath] is what I entered.
The pro addition for students is $60 US for one year.
It is well worth that price. Your school or university may have a site-license.
 
pka said:
First of all that is an online program Wolfram Alpha SEE HERE

[imath]3^{y^2+3}=x^2(a+b)^x[/imath] is what I entered.
The pro addition for students is $60 US for one year.
It is well worth that price. Your school or university may have a site-license.
Click to expand...
That means you entered the solution and not the problem.
 
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