Sorry. I was editing the post. It was not completed.

See now

Hi,

which step you didnt get ??

4th line RHS. How did you get 5th line RHS from 4th line RHS?

it is an exponent tule

\(2^{6x} - 2^6\)

Same base, then you can subtract the exponents

\(2^{6x-6}\)

shouldnt the x next to the 2 on the LHS?

Because the way I am reading the first line, is 2 to the power of 2 to the power of 3 x, and when you power it.

it would be 2 to the power of 8 to the power of x?


I dont understand why the x was carried over to the higher power than brought down?

It is not brought down.

it is a number raised to the power and this raised to the power.

it is 2 raised to the two and the latter raised to 3x

\({2^2}^{3x}\)

Then you multiply the exponents and the result is \(2^{6x}\)

I guess, I dont know where the 8 went?

In the answer above, is the LHS or the RHS the \(2^{8x}\)\(=640000\)?


edit:
In the problem, it is 8x, and x is multiplied by 8. I understand that \(2^{8}\) is like saying \({2^2}^{3}\). I am having a hard time getting why the x is next to the three, not the second two.


I also understand if the \({2^{{(2x)}^{3}}\), then the x will be cubed

I mean,
How did you get this \(5^4\) from this \(\frac{5^4}{2^3}\)?
or from \(\frac{2^{6x-6}}{2^3}= \frac{5^4}{2^3}\) to \(2^{2x-2} = 5^4\)? Need overall explanation because it not easy for me to understand.

Yes you are right. I made a mistake to think it was raised to x and not multiplied by x

Ok attack from another perspective

1) \(2^{8x} = 2^6 \times 10^4\)

2) \(2^{8x} = 2^6 2^4 \times 5^4\)

OR

\(\frac{2^{8x}}{2^8}= \frac{2^8 \times 2^2 \times 5^4}{2^8}\)

\(2^{8x-8} = 2^2 \times 5^4\)

Now clearly all the exponents can be divided by four

\(2^{2x-2} = 2^{\frac{1}{2}} \times 5\)

\(2^{2x-2} = 5 \sqrt{2}\)

Fully understand, Thanks.

Can we divide the exponents by 4 irrespective of the base number?

Please can you provide an explanation. ( I have never came acrossed this rule )

Exactly. That was my question too.

The final equation stands like this..
\(2^{8x-10} = 5^{4}\)

First of all the exponent is 8x-8 and not 8x-10

Secondly, I went down without some obvious step but indeed yes.

It is a simplification/manipulation.

regards

I had been waiting for you Sir and your insight.

Thank you

Got that,

Just trying out to simplify

\(2^{8x} = 640000\)

or \(2^{8x} = 2^6 * 2^4 * 5^4 = 2^{10} * 5^4\) ( 640000 = we know, 2^6 = 64 and now there are 4 zeros after 64, that means it will compiled with = 2^4 * 5^4 , since 2*5 =10)

Now taking square root on both sides

\(2^{4x} = 2^5 * 5^2\)

Dividing by \(2^4\)on both sides

\(\frac{2^{4x}}{{2^4}} = \frac{{2^5*5^2}}{{2^4}}\)

\(\frac{2^{4x}}{{2^4}}= 2 * 5^2\)

Taking square root again,

\(\frac{2^{2x}}{{2^2}} = 5 \sqrt{2}\)

or \(2^{2x - 2} = 5 \sqrt{2}\)

Dude. I am just quoting what you said before.

Now clearly all the exponents can be divided by four


\(2^{8x-8} = 2^{2} * 5^{4}\)

Dividing by 4 will give you

\(2 ^{8x-10} = 5^{4}\)

Am I missing anything?