Math Is Fun Forum

I don't seem to understand the step 3.

I thought, it would be something like:
xlog10 - log2 + xlog3 = 0

I claim the above because of the following:

2 - 2log5 = log10^2 - log5^2 = log4.

Why didnt it take xlog10. because 2 is the log and not the base. I think it cant be xlog2.

Please help me understand this.

Thanks in advance

Hi, bobbym I have come across another confusing one with different bases.

Solve for X.

3^x * 2^x-1 = 1.

How would one go about this? Is wierd.

Okay I digest it. But can I  use that procedure to arrive on that answer 7 you had in the previous problem?

Errrr okay I have got you, then what would one  make of the two, I mean the base.

I don't understand yours, I thought it would be

2*3*3(2n+3) = 2*9^(2n+3) = 2*3^3(2n+3) = 2*3^(6n+9)

It is due to such situations why I could not solve the previous problem since it has 6*3^2n+3. the six made it hard for me to manipulate. One cannot reduce it to be three. Or if the six had been three I would multiply it with the three to get nine and then reduce it to three, so I could manipulate. Since nine would be a perfect square.

But is it possible to add the exponents of the following? 

I have learnt that 5^0 = 1. Can you explain to me why it is equal to one?

Then I need to learn the laws of exponents, or do indices also teaches that? If not then please could you assist me learn it?

Thanks for your assistance, God bless!

Yes to be > 18^(3n+3). But I see the eighteen cannot further be reduced to three, in order to have the same base with the others.

But I don't seem to understand those methods