This doesnt make sense Log base 9 of(1/81) solving exponentially you get  -2 on a sci calc i get .01178 why?
914213
OK. It looks like you took the log base 10 of 9 (which is .9542) and multiplied it by 1/81 (which is .01234) to get your .01178. Thus, you found\n(1/81)*Log(9).\n\nTo get log base 9 of 1/81, i.e.\nlog_9 (1/81)\nyou have to find the exponent x where\n9^x=1/81.\nThis exponent will be -2 since\n9^(-2)=1/9^2=1/81.\nNo calculator is needed here!\n\nTo find the result on a calculator, you have to find\nLog(1/81) (which is -1.908485)\nand divide it by\nLog(9) (which is .9542425)\nto get -2.
140923
The other property of logarithms is that Log_x A - Log_x B = Log_x (A/B)\n\n\nso you can combine this rule with the other one i told you about in your other question...\n\nthe only thing is that u are working in reverse order\n\n\nLn 2y/x= Ln 2+ Ln y - Ln x\n\n\nThese are properties, they should be tought to you as principals, without them you cant solve logs:D hope this helps ;) if you think it does pls gimme 10 pts:P
716399
log x-log(x-3)= -1\nlog(x/(x-3))=-1\nx/(x-3)=10^-1=.1\nx=.1(x-3)=.1x-.3\n.9x=-.3\nx=-1/3
968058
Type 1.16 log / 4 / 1.0125 log =\nThis is 3 approximately (2.9869...)\n\nTh.
702351
I think you are looking too much into it, but here we go. \n\nIf you have the equation x^2 and say we pick points x= -1, and 1. \n\nWhen you plug the x's into the equation you will get a positive answer. This is why whenever the exponent is even, it will always bounce off the x- axis, because there is no way for it to go negative.\n\nfor the second question, say we take x^2 and plug in the same points above.\n\n-1 ^ 3 gives you -1. This would mean that ever x = -# will have a negative y value which means it will cross the x- axis\n\nTry it with more equations and you will understand
640099
Log2e^3=4x\nAssuming this means: log(2e^3)=4x and the log is to base 'e'\nlog(2e^3)=4x\n=> log(2) + log(e^3) = 4x\n=> log(2) + 3log(e) = 4x\n=> .693 + 3 = 4x   [half life formula? ln(2)=0.693, approx.]\n=> 4x = 3.693 [rounded approx.]\n=> x = 0.923 [rounded approx.]\n\nDid I just do some kid's homework?
1075997
Since you use log rather than ln, I assuming base 10 (common) logs.\n\nLogrithms are exponents!!!!!\nThey are the exponent you hang on the base to get the number in question.  That is,  saying   log x = y is the same as saying 10^y = x.\n\nSo log x = -2 is the same as saying 10^(-2) = x  or x = 1/100.
743256
First 4*log[x-2] = log[x-2]^4. \nThen -log[5] = log[1/5]\n\nThe sum of logs is the log of the product of the arguments and so this gives\n\n(1/2)*log[(x-2)^4 * (x+3) / 5]\n\nFinally the (1/2) becomes an exponent in the argument:\n\nlog[((x-2)^4 * (x+3) / 5)^(1/2)]
290613
The sum of the logs is the log of the products, ie.\nlog(a)+log(b)=log(ab)\n\nSo combine the sum of logs here into a single log of the product of (x-3) and (x+3).\nlog_4(x-3)+log_4(x+3)\n  =log_4((x-3)(x+3))=2\n\nOnce you have a single log isolated on one side of the euqation you should remember that the log equation log_a(b)=x is equivalent to the exponential equation a^x=b, so in this case\n(x-3)(x+3)=4^2=16\n\nNow you've got a quadratic equation that you can solve for x by expanding everything, moving everything to one side of the equation, and factoring.\nx^2-6x+9=16\nx^2-6x-5=0\n(x-5)(x-1)=0\nx=1 or 5\n\nYou should then check to see if both of these solutions are valid in the original equation. Since letting x=1 would involve taking the log of a negative number in the original equation (which you cannot do if you are dealing with real numbers) you should throw that solution out.\n\nx=5
350908
no this is how to work it out. I am assuming you know how to take the logarithm of a coplx number and find the modulus and argand of it.\n\ni^i = e^(logi^i) = e(ilogi)\n\nnow log i = log|i| + i(argi)\n= 0 + iπ/2 + i2kπ\n\nwhere k is an integer\n\nso substituting this is\ni^i = e^(i(iπ/2 + 2ikπ)\n= e^(-π/2-2kπ)]\n= e^(π[-1/2-2k])
739921
Not too hard.\n1. plug in what you know\n\nP/2 = Pe^(-0.0027x)\n\n2. cancel the Ps.\n1/2 = e^(-0.0027x)\n\n3. use some tricks with the ln and e to get x by itself\nln(1/2) = -0.0027x\n\n4. solve for x\nln(1/2) / (-0.0027) = x\n\n5. calculator, most definitely\nx = 256.72\n\n\nOn step 3, try to remember this:\na = e^b, therefore ln(a) = b. That's usually the make-or-break with exponential decays.
1181156
It looks like that first 1 should be an i. In this case,\ni^z=exp(i*pi/2*z),\nwhere I am using the principal branch of the logarithm to define the exponential.\nThen the derivative will be \n(i*pi/2)*exp(i*pi/2*z)=\npi/2*i^(z+1)\n\nEasy.
808044
Take the log of both sides:\nlog(0.01) = log(2.2^-t/8)\n\nRemember that log(a^b) = b * log(a), so bring the -t/8 in front:\nlog(0.01) = -t/8 * log(2.2)\n\nNow just solve algebraically:\n-t/8 = log(0.01) / log(2.2)\nt = -8 * log(0.01) / log(2.2)\n\nlog(0.01) = -2\nlog(2.2) ≈ 0.342422681\n\nt = (-8 * -2) / 0.0.342422681\nt = 16 / 0.342422681\n\nTherefore:\nt ≈ 46.7258768\n\nAs a double check, you can plug it in your calculator:\n2.2^(-46.7258768/8) = 0.01
807068
(-1.6875) (81) (0.28125) = 38.443359375\n\nIf thats not right, sue me.
494525
Think of it in terms of logarithms, since it is the inverse of exponents.\n\nStatement to prove:  1^infinity = 1\n\nRewritten in logarthmic form:  log[1](1) = infinity\n(Statement above is read: log base 1 of 1 equals infinity)\n\nWell that can't make sense, right?  It is a law of logs and of exponents that log[b](1) = 0 for any positive base b.  This is the same as saying b^0 = 1, for any b.  Also, log[b](b) = 1, for any positive b.  This makes since in b^1 = b.  Both of these laws would conflict this reasoning.  In fact, when b = 1, they conflict with one another.  Therefore, lets look at it with a variable instead.\n\nProve:  1^infinity = X\n\nRewrite in logartithmic form: log[1](X) = infinity\n\nNow, what value of X will give you infinity?\nCan you determine that?\n\nI would say it is indeterminate :)
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
840849
1.52582219663167... if the log is base e\n1.59874065990736... if the log is base 10\n\nAs the use of a graphing calculator does not give any insight into the derivation of the solution, here it is:\n\nTo solve this, you need to use some numerical interpolation method, as the equation can not be solved directly for x (in terms of finitely many algebraic functions).\n\nNewton's method yields:\nLet y = f(x) = log(x) + 3x - 5 \n{here log(x) is the natural log(base e).  We want to find the value of x for which f(x) = 0}\nf '(x) = 1/x + 3\nthe method says that x(n+1) = x(n) + (y - f(x(n)) / f '(x(n))\nand here, y = 0, so\n\nx(n+1) = x(n) + (5 - log(x(n)) - 3x(n)) / (1/x(n) + 3)\n\nthen, x = lim(n→∞) x(n)\n\n(You must choose something reasonably close for x(1), such as\nx(1) = 1.  It takes only five iterations to get the accuracy of my solutions above in a VB6 program using double precision variables.  i.e., x(5)=1.52582219663167)