Sorcery of Logarithms (Urgent)

Look at your equation....we can do this.....

3x + 3 = 10

3(x+1)  = 10     divide both sides by 3

x + 1 = 10/3     subtract 1 from both sides

x = 10/3 - 1   =    10/3 - 3/3  = 7/3

Now....without factoring we get

3x + 3  = 10   subtract 3 from both sides

3x = 7   divide both sides by 3

x = 7/3     ...exactly the same answer we got with factoring out the "3"  !!!

Notice that factoring does not involve adding or subtracting a quantity from both sides.....it's just a method used to simplify one side.... (or possibly, both sides).......

10^x - 0.2 * 10^x = 40

So we have, factoring

10^x(1 - .2)  = 40    simplify

10^x * (.8)  = 40   

10^x *(8/10) = 40   multiply both sides by 10/8

10^x  = 50              

This says that

log 50  = x  = about 1.698  or 1.7  if  we wish to round

Thank you for replying.

Can you explain how and why 1 is subtracted with 0.2?

10^x - 0.2 * 10^x = 40

Notice that 10^x is common to both of the first two terms.....so we can factor it out....so we have

10^x ( 1  -   .2)

Notice....that if we distributed the 10^x back......we would have what we started with....!!!

Does that help???

Sorry. You'll have to explain this on a retard level.

I don't understand how the subtraction with one happens and why it turns into 0.02 instead of 0.2.

Sorry...that was a mistype.....it should have been  .2

See if this helps

We really have  

(1) * 10^x - (.2) *10^x

What term is common to both......???.....answer....... the 10^x

So, factor this out and we have

10^x ( 1 - .2)   ....  and notice that 1 - .2  = .8

So we have

10^x (.8)

I understand the result and the last part. But what do you mean by (1)? What rules allow this 1 to come from nowhere and what rule allows 10^x to be removed because there is another 10^x? Maybe I lack the basics or something?

Notice that we can multiply anything by 1

So

(1)* 10^x  = 10^x

So

10^x - (.2)*10^x   is the same as

(1)* 10^x  - (.2) 10^x  

So.....we can factor out the 10^x   and we're left with

10^x ( 1 - .2)

Does that help??

I understand. But how do you remove the 10^x. I understand you are factoring out, but how doesn't this break against equation laws? Everything you change on < side you have to change on > side. For example

3x + 3 = 10

3x = 7

I understand this example perfectly clear. But I'm having some problem factoring on logarithms.

I know. Can you give me one very similiar to this one and I'll see if I can solve it.

It works just the same way, Vraces....we just factored out 10^x   rather than a "whole" number like 3....!!!

OK....try this one

10^x - .8* 10^x = 20

{you should get x= 2 for your answer}

10^x - 0.8 * 10^x = 20

10^2x + 0.2 = 20

10^2x = 4

10^x = 2

I still don't understand how + 0.2 can be multiplied to the other side ( 20 ).

Surely I can get the answer, but if I don't learn the purpose of w*f I'm doing, then I will not be able to mathematically evolve.

Thank you for your effort, and I'm sorry to take so much of your time.

OK, Vraces...let's look at this one...we have....

10^x - .8* 10^x = 20          note......that just like the other one, 10^x  is common to both of the first two terms......so ....factoring it out, we have

10^x * ( 1 - .8)  = 20

10^x  * (.2) = 20      divide both sides by .2

10^x = 100          and this says that

log 100 = x   = 2

What do you mean by " it is common to both" I am learning math in a Swedish school. And also, what would I do if there were for example 10^X -0.2 *10^x *10^x

10^x - .8* 10^x = 20 

"Common to both" means that the term 10^x  is found in both the first and second terms

If we had

10^x -0.2 *10^x *10^x  we could factor out 10^x and be left with

10^x (1 - .2 *10^x)

Distribute the 10^x across both these terms and you will have what we started with.....

I think I almost got it. Can you just explain why this method is wrong.

10^x - 0.8 * 10^x = 20

10^2x (1-0.8) = 20

10^2x + 0.2 = 20

10^2x + 2 = 200

10^2x = 100

10^x = 50

____________

10^x(1-0.8) = 20

10^x + 0.2 = 20

10^x + 2 = 200

10x^2 = 100

x = 2

+++Question: Why does can't I take -2 on both sides instead of dividing by two  on both sides? I'm asking these questions to understand better.

10^x(1-0.8) = 20

Note that we are multiplying  {not adding}

10^x  times (1 -.8)

And ( 1 - .8)  = .2

So we have

10^x *(.2)  = 40      divide both sides by .2

10^x  = 100  

but 40/0,2 is 200?

Sorry....I am trying to do too many things at once..

I meant to type

10^x *(.2)  = 20

So

10^x = 20/.2

10^x = 100

Ok. I pretty much got it covered now. But can you explain the significance of 10^x. Since you're taking it of the equation. How can this be. I know you probably already explained this 100 times.

Note that if we have

a^x  = b      we can write this in log form as

log b/ log a = x

So

10^x = 100

In log form, we're saying

(log 100) / (log 10) = x     but since log 10 = 1, we can just write this as

(log100) / 1   = log 100 = x   

And

log 100  = 2

So basically you "convert" one 10^x to 1 because log 10 = 1 ?

I'm not converting  10^x to 1

I'm writing 

10^x  = 100  in log form

If I take the log of both sides, we get

log10^x  = log 100        and by a log rule ....  loga^x =  x log a

So

log10^x   = x log 10.....so we have

x log 10  = log100    divide both sides by log 10

x = log 100 / log 10       but log 10 = 1    so we have

x =( log 100) / 1  = log 100  = 2

I was reffering to the

10^x -0.2 * 10^x

becoming just

(1 -0.2)10^x

I want to understand how this one of the 10^x disappears.

10^x (1 - .2 *10^x)

If I were to distribute the "10^x" over both of the terms  in the parentheses, we would have

10^x - .2 *10^x *10^x  ....just what we started with....

10^x - .2 *10^x *10^x

but we started with

10^x - .2 *10^x?

If we're talking about

10^x -0.2 * 10^x

Factor out the 10^x

10^x(1 - .2)

Note...if we distribute the "10^x" across both terms, we have what we started with

10^x - .2*10^x

What happens exactly when you factor out 10^x.

I don't understand your question.....we're just taking out the greatest common factor....just like normal Algebra

I don't know how to explain my problem.

I don't understand how

10^x - .2 *10^x

Transforms into 10^x(1 -.2)

The two problems I have are understand how 10^x dissappears and how the equation changes into (1-0.2)

What if we had this....???

3 - .2 (3)

Wouldn't this factor as

3(1-.2)  ???

Because.....distributing the 3 back across the parentheses would give us

3 - .2(3)    ....just what we started with....

So...it works the same with 10^x

10^x - .2 (10^x)  =

10^x( 1 - .2)

I finally got it. Thanks a lot. Sorry it took like literally 2 or more hours lel.

Hi Vraces - it is good to see you on the forum again :)

Thanks CPhill

I have gone through and given you both points for all your questions and answers.

Both were great.  We want people to be persistent around here.    

Vraces, I think that you may have a problem with factorising. 

If you want help with this you have only to log back on and ask for it. 

I'll show you some examples.

Please try to see that each time I have taken the common factor out the front and left the rest in the brackets.

$$\\\textcolor[rgb]{0,0,1}{6}-0.2*\textcolor[rgb]{0,0,1}{6}\\
=1*\textcolor[rgb]{0,0,1}{6}-0.2*\textcolor[rgb]{0,0,1}{6}\\
=\textcolor[rgb]{0,0,1}{6}(1-0.2)\\
=\textcolor[rgb]{0,0,1}{6}*0.8 \\
= 0.8*\textcolor[rgb]{0,0,1}{6}\\\\\\\\
\textcolor[rgb]{0,0,1}{y}-0.2\textcolor[rgb]{0,0,1}{y}\\
= 1*\textcolor[rgb]{0,0,1}{y}-0.2*\textcolor[rgb]{0,0,1}{y}\\
= \textcolor[rgb]{0,0,1}{y}(1-0.2)\\
=\textcolor[rgb]{0,0,1}{y}*0.8\\
=0.8\textcolor[rgb]{0,0,1}{y}\\\\\\\\
\textcolor[rgb]{0,0,1}{10^x}-0.2*\textcolor[rgb]{0,0,1}{10^x}\\
=1*\textcolor[rgb]{0,0,1}{10^x}-0.2*\textcolor[rgb]{0,0,1}{10^x}\\
=\textcolor[rgb]{0,0,1}{\textcolor[rgb]{0,0,1}{10^x}}(1-0.2)\\
=\textcolor[rgb]{0,0,1}{\textcolor[rgb]{0,0,1}{10^x}}*0.8 \\
= 0.8*\textcolor[rgb]{0,0,1}{10^x}$$