Wait! Note: do not use the graphical method...Any other method is acceptable...

Although the answer is quite obvious, the process of getting to the answer is almost nearly im-possible....

I would be sooo happy if this is put on the first page..

It has to do with finding the log... Gimme a minute. Been a while since I've done this shii

I think...8x=2(2)x 2×2)x=4x 8x÷4x=2. Just saying, dunno if it's right

8x=2^(2x), using log on both side log8+logx=(2x)log2, log2^3+logx=(2x)log2, logx=(2x)log2-3log2, then factorise log2 out logx=log2{(2x-3)}, log eliminate log remain x=2(2x-3), then x=4x-6, x-4x=-6, -3x=-6, x=2.

Horluwasegun:
8x=2^(2x), using log on both side log8+logx=(2x)log2, log2^3+logx=(2x)log2, logx=(2x)log2-3log2, then factorise log2 out logx=log2{(2x-3)}, log eliminate log remain x=2(2x-3), then x=4x-6, x-4x=-6, -3x=-6, x=2.

This method is completely wrong. This is called Bobo mathematics.... It is very impossible...I still won't mind if this could be put on first page.

rap53:


This method is completely wrong. This is called Bobo mathematics.... It is very impossible...I still won't mind if this could be put on first page.

Brother he is correct, the log is in base10, so factoring out is correct.

Tiny23:
I think...8x=2(2)x
2×2)x=4x
8x÷4x=2.
Just saying, dunno if it's right

My God!! See illegal manipulation which mathematical method did you used in bringing down the power 2x

lexiconkabir:


My God!! See illegal manipulation which mathematical method did you used in bringing down the power 2x

lol

lexiconkabir:


Brother he is correct, the log is in base10, so factoring out is correct.

No, he isnt right!

He missed it from here..

factorise log2 out logx=log2{(2x-3)}, log eliminate log remain x=2(2x-3), then x=4x-6, x-4x=-6, -3x=-6, x=2.


Kindly review your logarithmic rules again.

rap53:


No, he isnt right!

He missed it from here..

factorise log2 out logx=log2{(2x-3)}, log eliminate log remain x=2(2x-3), then x=4x-6, x-4x=-6, -3x=-6, x=2.


Kindly review your logarithmic rules again.

Ok, that log what is its base by default? Or what exactly do you see as the problem there? Cancellation of the log or factorization of the log? But if you still don't get it, instead of taking natural log of both sides in the beginning, take the ln of both sides (log to base e).

lexiconkabir:


Ok, that log what is its base by default? Or what exactly do you see as the problem there? Cancellation of the log or factorization of the log? But if you still don't get it, instead of taking natural log of both sides in the beginning, take the ln of both sides (log to base e).

The thing exactly wrong there is the removal of the natural log on both sides and then 2(2x-3),multiplying like that on the right hand side. There is no rule in logarithm that explains this. You are free to check any mathematics textbook in this world to confirm this

lexiconkabir:


Ok, that log what is its base by default? Or what exactly do you see as the problem there? Cancellation of the log or factorization of the log? But if you still don't get it, instead of taking natural log of both sides in the beginning, take the ln of both sides (log to base e).

The thing exactly wrong there is the removal of the natural log on both sides and then 2(2x-3),multiplying like that on the right hand side. There is no rule in logarithm that explains this. You are free to check any mathematics textbook in this world to confirm this.

.

@rap53 ok let me stop beating around the bush,

From logx=log2(2x-3) right?

We have logx - log2(2x-3) = 0

Log{x/2(2x-3)} = 0

x/2(2x-3)=10^0 → x/2(2x-3)=1

X=2(2x-3) and it follows from there....

So this are the steps that were omitted, which implies canceling directly is correct.

first you divide through by 2

you get 4x =2x(2x came down)

you divide through by 2 again it gives.... 2x=x

and we all know x equals 1 sometimes

so 2x becomes 2

hence X=2....you are welcome

lexiconkabir:
@rap53 ok let me stop beating around the bush,

From logx=log2(2x-3) right?

We have logx - log2(2x-3) = 0

Log{x/2(2x-3)} = 0

x/2(2x-3)=10^0 → x/2(2x-3)=1

X=2(2x-3) and it follows from there....

So this are the steps that were omitted, which implies canceling directly is correct.

Actually, you should know that the 2x-3 is multiplying log 2. What you did was put 2x-3 to multiply 2 in the bracket. whereby, the RHS of the equation can only be written as 2x-3log2 or log2^2x-3. Hence, my issue with the equation.