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Mathematics 7 Online
mathslover (mathslover):

Logarithm Question Find x, if \(\cfrac{6}{5} a^{\log_a x \log_{10} a \log_a 5 } - 3^{\log_{10} (x/10)} = 9^{\log_{100} x + \log_4 {2}} \)

mathslover (mathslover):

\(\LARGE{\cfrac{6}{5} a^{\log_a x \log_{10} a \log_a 5 } - 3^{\log_{10} (x/10)} = 9^{\log_{100} x + \log_4 {2}}}\)

mathslover (mathslover):

@hartnn @ganeshie8

hartnn (hartnn):

those are lots of logs :P

mathslover (mathslover):

Haha :D

mathslover (mathslover):

Should I simplify RHS first?

hartnn (hartnn):

easiest simplification i see is for log_4 2

mathslover (mathslover):

Haha.. :D But, RHS becomes : \(\LARGE 3^{\log_{10} x + 1}\)

ganeshie8 (ganeshie8):

maybe use this repeatedly for LHS : mlogn = logn^m

hartnn (hartnn):

left side , 1st term exponent! simplifies greatly

mathslover (mathslover):

Yeah..

mathslover (mathslover):

doing it.. \(\LARGE \cfrac{6}{5} x^{\log_{10} a \log_a {5}}\)

mathslover (mathslover):

the first term in LHS ^

mathslover (mathslover):

\(\LARGE \cfrac{6}{5} x^{\cfrac{1}{\log _ a{10} } \times \log_a 5}\) hmm, no:/

mathslover (mathslover):

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

ganeshie8 (ganeshie8):

\(\large a^{\log_a x \log_{10} a \log_a 5 } \text{ simplifies to } 5^{\log_{10} x}\)

hartnn (hartnn):

use change of base repeatedly

mathslover (mathslover):

or shall I do like this : \(\LARGE \cfrac{6}{5} x^{\cfrac{\log_x a}{\log_x {10}} \log_a 5}\) \(\LARGE \cfrac{6}{5} a^{\cfrac{\log_a 5}{\log_x {10}}}\) \(\LARGE \cfrac{6}{5} 5^ {\cfrac{1}{\log_x {10}}}\)

mathslover (mathslover):

or \(\LARGE \cfrac{6}{5} 5^{\log_{10} x}\) o.O finally got a short expression.

mathslover (mathslover):

Okay, so , it becomes : \(\LARGE \cfrac{6}{5} 5^{\log_{10} x} = 3^{\log_{10} (x/10)} + 3^{\log_{10} x + 1}\)

mathslover (mathslover):

\(\LARGE 6 \times 5^{\log_{10} (x-1) } = 3^{\log_{10} x } \times \left(\cfrac{1}{3} + 3\right) \)

mathslover (mathslover):

I think, we got it : if I keep it 6/5 then : \(\LARGE {\cfrac{6}{5} \times 5^{\log_{10} x } = 3^{\log_{10} x } \times \cfrac{10}{3} \\ \left(\cfrac{3}{5} \right)^2 = \left(\cfrac{3}{5}\right) ^{\log_{10} x}} \) \(\boxed{\bf{x = 100}} \ddot \smile\)

mathslover (mathslover):

Got it.. Thanks @ganeshie8 and @hartnn .

ganeshie8 (ganeshie8):

Wow ! lol im still wasting my paper... looks great :)

mathslover (mathslover):

Haha...! Purchase some more for me... more to come. :)

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