If b2(square)=ac Prove That 2/logbx=1/logax+1/logcx?
924703
b^2=ac\nb^2x^2=ax*cx\n(bx)*(bx)=ax*cx\nflipping\n1/(bx)*1/bx=1/ax*1/cx\ntaking log\nlog1/bx+log1/bx=1/logax+1/logcx\n2/logbx=1/ax+1/cx
797419
Let y = ax² + bx + c\n\nWhen x = 1; y = a + b + c = 12 ............ (1)\nWhen x = 2; y = 4a + 2b + c = 22 ........ (2)\nWhen x = 3; y = 9a + 3b + c = 30 ........ (3)\n\nOn solving you get a = -1 b = 13 c = 0\n\nie y = 13x - x²\n\nAgree with Antoine above
477145
log(2x+1) - log5  = log x\nlog(2x+1) - log x = log 5\nlog((2x+1)/x)  = log 5\n(2x+1)/x  = 5\n2x+1 = 5x\n3x = 1\nx= 1/3
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
708644
Use the quadratic formula.\nx1 = 1\nx2 = -1/2
46798
if (1/x) + 2= (c/2)\n--------------------------------\n\n1/x = (c/2) - 2\n----------------------\nx = 1/((c/2) - 2)\n       (OR)\nx = ((c/2)-2)^-1\n-----------------------
627488
you can change the base...\nln x = (log2 (x))(ln 2)
783096
yes, it really is easier than you think:\na^(log_a(2/x) +log_a(x))=a^(log_a(2) -log_a(x) +log_a(x))\n  =a^(log_a(2))\n  =2\nnote that i used the property log(a/b)= log(a)- log(b) and a^log_a(x)=x
883916
Let log_2(x) denote the log of x with base 2.\n\nUse the identity that log_2(x) = ln(x) / ln(2). This identity holds because:\n\nRecall that, by definition, 2^(log_2(x)) = x.\n\nln(x) = ln(2^(log_2(x))) = (log_2(x)) (ln(2)), so dividing through by ln(2) shows the result.
100388
This is due to a property of log which is that :\nlog(ab)=log(a)+log(b)\nRemark that this is true when well defined (for eg you can't write log(-1*-3)=log(-1)+log(-3) )\n\nSo here, you have log(x^2)=log(x)+log(x)\n\nTherefore, there are no diff. except that log(x^2) exists for all x in R&#92;{0} and 2log(x) is defined for all x strictly positive.
941613
log_a b = x is the number to which you need to raise a to get b:\na^x =b\n...........
911082
logx+log(x+48)=2\n=>log{x(x+48)}=2\n=>log(x^2+48x)=2\n=>x^2+48x=10^2\n=>x^2+48x-100=0\n=>x^2+50x-2x-100=0\n=>x(x+50)-2(x+50)=0\n=>(x+50)(x-2)=0\nhence,\nx=-50 or x=2
329355
Try solving this one:\n\nDetermine all pairs of positive integers (a, b) such that\n\na^2/(2ab^2-b^3+1)\n\nis a positive integer.\n\nNote that in 2ab^2 just the b is squared.\n\nGood luck!
541208
it is a proving isn't it?\nright handside: [logbase10 1000]/logbase10 2\n=[logbase10 10^3]/[(logbase2 2 )/( logbase2 10)]\n=3[logbase10 10]x[logbase2 10]\n=3logbase2 10\n=logbase2 1000 , proved\n\n\n\nif its right do mark it best i'll come to know i have done correct or not.\nsorry here is the right answer\nanswer is 4\ni got it by\nlogbase2 1000=1ogbase10 1000 / logbase100 "x"\n3logbase2 10=31ogbase10 10 / logbase100 "x"\nlogbase2 10=1/logbase100 "x"\nlogbase100 10 / logbase100 2 =1/logbase100 "x"\n.5logbase100 100 / logbase100 2 =1/logbase100 "x"\n2logbase100 2=logbase100 "x"\nlogbase100 2^2=logbase100 "x"\n2^2="x"\n"x"=4
402319
ya it is a^2 +b ^2 +2abcos(c)
750081
log (ab)= log a + log b\nlog (a^p) = p log (a)\nlog (a/b) = log a - log b !
975217
Line AC = 6, line BC= 2x+2, line AB=x+1\nSo 2x + 2 = 6\n2x = 4\nx = 2 <-- option C