My 15th Problem

Algebra Level 4

$\large { 2 }^{ 2 (\log _{ 10 }{ 4x}) -1 }-{ 7 }^{ \log _{ 10 }{ 4x } }={ 7 }^{ (\log _{ 10 }{ 4x} ) -1 }-3 \times { 4 }^{ \log _{ 10 }{ 4x } }$

What is the value of $x$ satisfying the equation above?

The answer is 25.00.

1 solution

Parth Lohomi
Mar 6, 2015

Let log(4x)=m

Then

$2^{2m-1}-7^m=7^{m-1}-3\times{2^{2m}}$

By solving (after taking log)

We get (real m)

$2^{2m}/2+3.2^{2m}=7^{m}/7+7^m$

$2^{2m}(7/2) = 7^m(8/7)$

$2^{2m}(49/16) = 7^m$

$\implies$ $4^{m-2} = 7^{m-2}$

$(m-2)\ log{4} = (m-2)\ log{7}$

The only solution is $m-2=0$

m=2

$\text{log}_{10} 4x = 2$

$4x=100\implies$ $\boxed{x=25}$

Nice solution.....

manish bhargao - 6 years, 3 months ago

thnx!! @manish bhargao

Parth Lohomi - 6 years, 3 months ago

did the same process..

Trishit Chandra - 6 years, 3 months ago

Thanks. I have updated the problem. It did not help that the problem was badly phrased and so many people ignored it.

@Utkarsh Bansal Please set a suitable level for your problem. This is not a level 5 problem, but more of a level 3 problem as it is a standard high school problem (once it has been clearly phrased).

Calvin Lin Staff - 6 years, 3 months ago

Sir I understood and thanks for your suggestion

Utkarsh Bansal - 6 years, 3 months ago

damn i always careless....i got m = 2 ..forgot sub back ....damn

Zack Yeung - 6 years ago

My 15th Problem

The answer is 25.00.

1 solution

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