Another Math Problem

[Zylios]

I don't know what it's called actually, just know you need to do logarithm.

I can work it out when it's 11^x=15, but then there comes one like this: 51^2x=2, that 2 before the x it fucks it up, I did it like this:

51^2x=2
log51^2x=log2
2x*log51=log2
2x=log2/log51=0,3010/1,7076=0,1763
x=0,1763/2=0,088

This is of course totally wrong somewhere, could someone tell me where? Thank you very much in advance

[Deleted]

51^2x=2
51^2x = (51^2)^x = 2601^x = 2
x = log2/log2601 = ~0,088

Edit;

Your solution is fine too, just plug x back in and you can check it (you just rounded down somewhere)

[Elim Garak]

it should be shorter

log51(2) = 2 * x, x = 0.5 * log51(2) = 0.088145717195

[Zylios]

I don't get it, could you try setting it up another way? Thank you so much anyway

[Elim Garak]

You need to find X. X is in the power -> it's a result of Log of 2 to base 51 -> 2x = Log of 2 to base 51 = 0.17629143439

Now just divide it by 2 to get pure X.

To put it more simply - you need to solve 51^x = 2, and then divide the result by 2

51^x = 2 -> x = log of 2 to base 51 -> log51(2) = log(2)/log(51) <- for calculator

51^2x = 2
log51(2) = 2x
log(2)/log(51) = 2x
x = 0.5 * log(2)/log(51) = 0.088145717195

[Zylios]

But it's 51^2x=2 not 51^x=2.

[Deleted]

You did it correct.

Pretend for a second 2x is another variable, let say y. and than calculate y (the same thing as if it was x, z or "blue"). than you have the equation y=2x so x =y/2 and since you calculated y, make it half is no problem.

Edit
For the record I think if you tested it in a calclator and got another answer than 2 or approximately 2 since you did approximations in you calculation. My guess is that you used the calculator wrong. Your calculation is correct.

[Elim Garak]

But it's 51^2x=2 not 51^x=2.

Remember what logarithm is:
The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number

number is 2
base is 51
exponent is 2x -> 2 * x

so calculate log51(2) = log(2)/log(51)
you will get the value of 2x -> 2 * x
so x is half of that.

[Zylios]

51^2x=2
log(51^2x)=log(2)
2x*log(51)=log(2)
2x*(1,7076)=log(2)
2x=log(2)
x=1,7076/2=0,8538

Is that correct?

[Deleted]

51^2x=2
log(51^2x)=log(2)
2x*log(51)=log(2)
2x*(1,7076)=log(2)
2x=log(2)
x=1,7076/2=0,8538

Is that correct?

Your FIRST calculation was correct!!!!! Why didnt you read what we replied?

[Zylios]

Because it can't be right. My teacher told me you could check if it was right by insetting the number you get in x's place, and then it gives me a number like 292435 so it's fucked

[Deleted]

Because it can't be right. My teacher told me you could check if it was right by insetting the number you get in x's place, and then it gives me a number like 292435 so it's fucked

Thats because you used your calculator wrong. Not because your calculation was wrong.

Edit
I used x=0.088 and got that 51^(2*0.088)=1.9977 or about 2.

Make sure you dont calculate (51^2)*0.088 = 229
or 51^(2/0.088)=6.4*10^38

[Deleted]

Then you did your check wrong; 51^(2*0.088) is indeed (rounded) 2.

[Elim Garak]

Because it can't be right. My teacher told me you could check if it was right by insetting the number you get in x's place, and then it gives me a number like 292435 so it's fucked

This is what I get in my calculator by inserting the result in place of x: 2.0000000000087908086233587369781
your teacher must be full of shit. Just use the shorter version, there's really no need for all those transformations you did. Maybe that's there your teacher got confused.

[Zylios]

Now I fucking get it! Thank you so much!