Solve Tricky Trigonometry Problems - Help Needed

  • Thread starter vijay123
  • Start date
  • Tags
    Trignometry
In summary, nazzardyes, the problem is trignometric, but all his efforts fail.and for info. this sum is a non calculator one...so...anyway..for the first question, it might be helpful to expand out 10 into 5*2(accoring to me). the brought the 2 to the other side and the after further simplying, found out that (16/9)^x = 240...i don't know how to go on without using a calc. hope some1 has a better solution to the problem.yes, i have tried all of them, they don't seem to work!
  • #36
see, here's how i did it...2^log2/3x^log2=(9x)^log3
((2^log2)/(3^log3))^(1/log6)=3x
2^log(6)2/3^log(6)3=3x
2/3=3x
x=2/9
 
Physics news on Phys.org
  • #37
sry mist some steps...but u ll get the picture once you do it yourself...you think the method is correct?...the ans. is correct coz i checked it in my graphing calc.
 
  • #38
Uhm, sorry for misreading the problem. But by the way, you should use parentheses to make the expressiona clearer. Saying:
b) (2/3x)^(log2) = (9x)^(log3)
is a bit confusing. Since, there are 2 ways to read it.
Is it:
[tex]\left( \frac{2}{3} x \right) ^ {\log 2} = (9x) ^ {\log 3}[/tex]
or:
[tex]\left( \frac{2}{3x} \right) ^ {\log 2} = (9x) ^ {\log 3}[/tex]
-------------
Your work looks good to me. The method is correct. Just missing a few steps. Congratulations. :)
By the way, do you know how to show that:
[tex]\frac{2 ^ {\log_6 2}}{3 ^ {\log_6 3}} = \frac{2}{3}[/tex]?
 
Last edited:
  • #39
yes...we have learned how to prove it...thanks for your help
 
  • #40
omg...how do you show that 2^(log2/log6)=2, by the case above...
i have only learned that 2^(log6/log2)=6!
 
  • #41
vijay123 said:
omg...how do you show that 2^(log2/log6)=2, by the case above...
No, this is not correct.
[tex]2 ^ {\frac{\log 2}{\log 6}} = 2 ^ {\log_6 2} \neq 2[/tex]!
---------------
In general, you can show that:
[tex]\frac{a ^ {\log_{ab}a}}{b ^ {\log_{ab}b}} = \frac{a}{b}[/tex].
Since the base of the log in both numerator and denominator is ab. So, to apply the property of inverse function: f(f-1(x)) = x, or f-1(f(x)) = x, i.e [tex](ab) ^ {\log_{ab} x} = x[/tex].
We should first multiplying both numerator, and denominator by either [tex]b ^ {\log_{ab}a}[/tex] or [tex]a ^ {\log_{ab}b}[/tex] to make it have the form above.
So:
[tex]\frac{a ^ {\log_{ab}a}}{b ^ {\log_{ab}b}} = \frac{\left( a ^ {\log_{ab}a} \right) \left( a ^ {\log_{ab}b} \right)}{\left( b ^ {\log_{ab}b} \right) \left( a ^ {\log_{ab}b} \right)} = \frac{a ^ {\log_{ab} a + \log_{ab} b}}{\left( ab \right) ^ {\log_{ab} b}} = ...[/tex]
Apply this to your problem, Note that: 6 = 2 . 3
Can you go from here? :)
 

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
906
  • Calculus and Beyond Homework Help
Replies
3
Views
948
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Replies
2
Views
1K
  • General Math
Replies
8
Views
961
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Precalculus Mathematics Homework Help
Replies
2
Views
4K
  • Calculus and Beyond Homework Help
Replies
4
Views
804
  • Calculus and Beyond Homework Help
Replies
4
Views
937
  • Precalculus Mathematics Homework Help
Replies
4
Views
1K
Back
Top