Logarithm Question Find x, if \(\cfrac{6}{5} a^{\log_a x \log_{10} a \log_a 5 } - 3^{\log_{10} (x/10)} = 9^{\log_{100} x + \log_4 {2}} \)
\(\LARGE{\cfrac{6}{5} a^{\log_a x \log_{10} a \log_a 5 } - 3^{\log_{10} (x/10)} = 9^{\log_{100} x + \log_4 {2}}}\)
@hartnn @ganeshie8
those are lots of logs :P
Haha :D
Should I simplify RHS first?
easiest simplification i see is for log_4 2
Haha.. :D But, RHS becomes : \(\LARGE 3^{\log_{10} x + 1}\)
maybe use this repeatedly for LHS : mlogn = logn^m
left side , 1st term exponent! simplifies greatly
Yeah..
doing it.. \(\LARGE \cfrac{6}{5} x^{\log_{10} a \log_a {5}}\)
the first term in LHS ^
\(\LARGE \cfrac{6}{5} x^{\cfrac{1}{\log _ a{10} } \times \log_a 5}\) hmm, no:/
Well, I will transpose that second term to RHS : \(\LARGE \cfrac{6}{5} x^{\log_{10} a \log_a {5}} = 3^{\log_{10} (x/10) } + 3^{\log_{10} x + 1}\) @hartnn from where did that come from? o.O
\(\large a^{\log_a x \log_{10} a \log_a 5 } \text{ simplifies to } 5^{\log_{10} x}\)
use change of base repeatedly
or shall I do like this : \(\LARGE \cfrac{6}{5} x^{\cfrac{\log_x a}{\log_x {10}} \log_a 5}\) \(\LARGE \cfrac{6}{5} a^{\cfrac{\log_a 5}{\log_x {10}}}\) \(\LARGE \cfrac{6}{5} 5^ {\cfrac{1}{\log_x {10}}}\)
or \(\LARGE \cfrac{6}{5} 5^{\log_{10} x}\) o.O finally got a short expression.
Okay, so , it becomes : \(\LARGE \cfrac{6}{5} 5^{\log_{10} x} = 3^{\log_{10} (x/10)} + 3^{\log_{10} x + 1}\)
\(\LARGE 6 \times 5^{\log_{10} (x-1) } = 3^{\log_{10} x } \times \left(\cfrac{1}{3} + 3\right) \)
I think, we got it : if I keep it 6/5 then : \(\LARGE {\cfrac{6}{5} \times 5^{\log_{10} x } = 3^{\log_{10} x } \times \cfrac{10}{3} \\ \left(\cfrac{3}{5} \right)^2 = \left(\cfrac{3}{5}\right) ^{\log_{10} x}} \) \(\boxed{\bf{x = 100}} \ddot \smile\)
Got it.. Thanks @ganeshie8 and @hartnn .
Wow ! lol im still wasting my paper... looks great :)
Haha...! Purchase some more for me... more to come. :)
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