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\(ln2^{4x-1}=ln8^{x+5}+log_2 16^{1-2x}\)

the right handside \(log_2 16^{1-2x}\) we can make the power (1-2x) next to the log so we could solve for 2^y=16 which will be 4 

\((1-2x)*log_2 16\)

\(log_2 16 = 4\)

4*(1-2x) + \(ln8^{x+5}\)=\(ln2^{4x-1}\)

4-8x+(x+5)*\(ln8\)=(4x-1)*\(ln2\)

\(\frac{4-8x+(x+5)*ln8}{4x-1}=ln2\)

Expressed in terms of ln2 

Further simplify could be done by using calculator 

ln8=2.07944154

4-8x+(x+5)*2.07944154

4-8x+2.079444154x+10.3972077

4-5.920x+10.3972077 

14.3972077-5.920x/4x-1 =ln2 

ln2=0.693147

14.4-5.9x=(4x-1)*0.693147

Rest is just algebra and indeed you will find that 

indeed you will find x=1.73589087 approx: 1.74 

Which works.

I don't think the question wants the x value so just expressed it in ln2 

asked and answered.

i think it just means moderator? :) 

oh wait! you were referring to something else LOL!

ok, no problem. do you know how I can get in touch with any of the moderators or do I just have to wait for them to see my question

thanks melody! :)

yup the next 3 week will be a tornado

Begin:o=0;e=0;listforeach(m, d=(10, 11, 12, 13, 14, 15, 16, ...........496, 497, 498, 499, 500),if(m%2==0,e=e+m, o=o+m);print"Even Num =",e; print"Odd Num   =", o

Even Numbers = 62,730
Odd Numbers   = 62,475

i am clacuser not lgged in

Evaluate the terms whenever one + + - - occurs.

For example.

1 + 2 - 3 - 4 = -4

5 + 6 - 7 - 8 = -4

Then find how many +  + - - s are there

3 and 4/9

Plug it in a calculator

I don't get it is 5304 the answer for the first one? and could you explain a bit more on how you got there?

a)     sum_(n=1)^26 (-1)^(n + 1) (105 - 4 n)^2 = 5304

for a)

I THINK you use the special case a²-b² = (a-b)(a+b)

never mind i got it

do you know the answer though? it's ok even if you can answer one

wait why does it say waiting for moderation?