The browser which is used for posting questions is almost totally useless for mathematical questions since it blocks most symbols.
I am assuming that your question is about log base 3 of (x plus 1) plus log base 2 of (x-1).
{log[(x + 1)^log2} + {log[(x - 1)^log3}/log(3^log2) where all the logs are to the same base - whichever you want. The denominator can also be written as log(3^log2)
This can be simplified (?) to log{[(x + 1)^log2*(x - 1)^log3}/log(3^log2).
As mentioned above, the expression can be to any base and so the expression becomes
in base 2: log{[(x + 1)*(x - 1)^log3}/log(3) and
in base 3: log{[(x + 1)^log2*(x - 1)}/log(2)
2x2+7/x1
logx +7=1+log(x-1) 6=log(x-1)-logx 6=log[(x-1)/x] 10^6=(x-1)/x 1,000,000x=x-1 999,999x=-1 x=-1/999,999
d/dx (2 log(1) + x) = 1
If log644 = x, then 64x = 4. The cubed root of 64 (which is the same as 641/3) is 4, so log base 64 of 4 is 1/3.
Log(3 * 1/3) = log(1) = 0
log(x)+log(8)=1 log(8x)=1 8x=e x=e/8 You're welcome. e is the irrational number 2.7....... Often log refers to base 10 and ln refers to base e, so the answer could be x=10/8
2x2+7/x1
Original Statement:x - 1 + 2 + log(x) = 3Simplify:x + 1 + log(x) = 3Subtract 1:x + log(x) = 2Lambert W-Function:x = (W(100*ln(10))/(ln(10)) = 1.7555794993... (rounded up).This considered log(x) to be base 10 log (x).
acording to me the value is 0 because the value of log 1 at any base is always 0.
log(x) + 4 - log(6) = 1 so log(x) + 4 + log(1/6) = 1 Take exponents to the base 10 and remember that 10log(x) = x: x * 104 * 1/6 = 10 x = 6/1000 or 0.006
Due to the rubbish browser that we are compelled to use, it is not possible to use any super or subscripts so here goes, with things spelled out in detail: log to base 2a of 2b = log to base a of 2b/log to base a of 2a = [(log to base a of 2) + (log to base a of b)] / [(log to base a of 2) + (log to base a of a)] = [(log to base a of 2) + (log to base a of b)] / [(log to base a of 2) + 1]
ln is the natural logarithm. That is it is defined as log base e. As we all know from school, log base 10 of 10 = 1 just as log base 3 of 3 = 1, so, likewise, log base e of e = 1 and 1.x = x. so we have ln y = x. Relace ln with log base e, and you should get y = ex
logx +7=1+log(x-1) 6=log(x-1)-logx 6=log[(x-1)/x] 10^6=(x-1)/x 1,000,000x=x-1 999,999x=-1 x=-1/999,999
- 1
It is zero
If the variables are x1 & x2 the solution is : 1) x1=x1+x2; 2) x2=x1-x2; 3) x1=x1-x2; EX: x1=1 , x2=6; 1) x1= 1+6 = 7 2) x2= 7-6 =1 3 x1=7-1 =6 ============================================
The derivative of ln x, the natural logarithm, is 1/x.Otherwise, given the identity logbx = log(x)/log(b), we know that the derivative of logbx = 1/(x*log b).ProofThe derivative of ln x follows quickly once we know that the derivative of ex is itself. Let y = ln x (we're interested in knowing dy/dx)Then ey = xDifferentiate both sides to get ey dy/dx = 1Substitute ey = x to get x dy/dx = 1, or dy/dx = 1/x.Differentiation of log (base 10) xlog (base 10) x= log (base e) x * log (base 10) ed/dx [ log (base 10) x ]= d/dx [ log (base e) x * log (base 10) e ]= [log(base 10) e] / x= 1 / x ln(10)