3^x=10.4\nHOw do I solve this? Thanks for any help.
208914
3^x = 10.4\n\nsince both sides are always positive, you can take the log of both sides...\n\nlog(3^x) = log(10.4)\nx*log(3) = log(10.4)\nx = log(10.4)/log(3)\n\nYou can use a calculator to figure out the log of 10.4 and 3.\n\nSteve
514818
So you have the formula, and you have the number to plug into it. Are you confused about how to deal with the exponents? \n\nIntensity = 5.4 x 10-10, so\n\nd= 10 * log( 5.4 x 10-10/ 10-12)\nd= 10 * log(5.4 / 10-2 )  \nd= 10 * log(540)\n\nd= 10 * 2.732                        note it is a base 10 log!!\nd= 27.32
593653
3x10^2 = 300 because 10^2 = 100\n3 x 10^(-1) = .3 because 10^(-1) = 1/10^1 = 1/10 = .1\n4 x 10^(-4) = .0004 because 10^(-4) = 1/10^4 = 1/10000 = .0001\n\nAll together it's 300.3004
185357
There is a rule of exponents that says (x^a)(x^b) = x(a+b).  Use this in reverse:\n\n(3^2x) - 3^(2x - 1) = 18\n(3^2x) - [3^(2x)*3^(- 1)] = 18\n(3^2x) - (3^(2x))/3 = 18\n3(3^2x) - (3^2x) = 54\n2(3^2x) = 54\n(3^2x) = 27\nlog(3^2x) = log27\n2xlog3 = log27\nx = log27/2log3\n= 1.5
229974
Are you sure it is x^3 and not x^2? If you're sure, then there is one relatively easy way to solve the equation:\n\n(x^3)-4x=24\n\nNow, this equation is of the form: x^3+px=q (where p=-4, q=24). The solution to this equation was given by the Italian mathematician "Cardan" to be:\n\nx=(sqrt((p^3)/27 + (q^2)/4) + q/2)^(1/3) - (sqrt((p^3)/27 + (q^2)/4) + q/2)^(1/3)\n\n\nFor your case, if you plug in the values for p,q you will get:\n\nx = (√(-64/27+144)+12)^(1/3) - (√(-64/27+144)-12)^(1/3)\n\nx = 2.888 - (-0.4629) = 3.3434\n\n\nThus, x=3.3434 is the real solution to the equation.
610926
it would actually be 4.3927  x  10 to the 3rd power.\n\nyou move the decimal over so that there is only one digit before it, and then you get that # and multiply it by 10 to (the amount of spaces the decimal moved to the left) power.\n\nif it is something such as 0.00043927, then it would be 4.3927  x  10 to the -4th power. when you move the decimal to the right, the power is negative.
540238
The second option is right.\n3x^(-2) = 1/3x^2
1391
Here's the best I can do. The answer I got is:\n\n24x^4 - 10x^3 + 690x^2 + 3190x = 1217
342814
Example:\n4^x = 8\n\nThe technique here is we change both sides into the same base.  Since we know that 4 = 2² and 8 = 2³, then we can change the equation into\n2^2x = 2^3\n\nThen, since the bases are equal, then the exponents must also be equal.  Thus,\n2x = 3\n\nTherefore,\nx = 3/2.\n\n----------------------------\nThe thing here is to change both sides into the same base.\nAnother example:\n81^[x(x + 2)] = 1/3^(9 - 20x)\n\nWe change both bases: 81 = 3^4, 1/3 = 3^-1.  Thus,\n3^[4x(x + 2)] = 3^[-(9 - 20x)]\n\nWe then equate the exponents:\n4x(x + 2) = (20x - 9)\n\nWe distribute:\n4x² + 8x = 20x - 9\n\nWe transpose everything to the left\n4x² - 12x + 9 = 0\n\nWe factor out\n(2x - 3)² = 0\n\nSquare root both sides\n2x - 3 = 0\n\nWe transpose -3\n2x = 3\n\nDivide 2.  Therefore,\nx = 3/2\n\n--------------------------\nHere's another example:\n\n2^x = 3\n\nI know that it is VERY DIFFICULT to change 3 into a power of 2.  Here, we use another tool: the logarithm.  If you are already ready, you can inform me so that I can teach you that.  Please feel free to contact me.\n\nkevin!\n^_^
427888
Using the quotient rule,\n\nlog(2x-1) - log(x-3)=1\nlog((2x-1)/(x-3)) = 1\n\nThen change to exponential form:\n\n10^1 = (2x-1)/(x-3)\n\nthen solve:\n\n10(x-3)=2x-1\n10x-30=2x-1\n8x=29\nx=29/8
167999
It will be C\nI'll explain you why\nsince function is 3^x, no matter which real number you put into the function, the result will be more than 0. If you put a negative number, you'll get a fraction. If you put a positive number, you'll get a power of 3.\n\nExample.\nx=-4\ny= 3^(-4)=1/3^(4)= 1/27\n\nwhich is a positive number, small, but positive.
676804
maybe something like 7^(1/3) over 7^(1/2)\n\nthat's the same as the cube root of 7 over the square root of 7.\nTo solve it, you would just subtract the exponents and you'd get 1 over 7^(1/6)\n\nI hope that is what you are looking for.
269949
Lets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3
661160
If you mean sqrt(x+4) = 3, then the answer is x = 5 since\n\nsqrt(x+4) = 3\nSo x+4 = sqrt(x+4)^2 = 3^2 = 9.\nSo x+4 = 9\nSo x = x+4-4 = 9-4 = 5.\n\nIf you mean sqrt(x) + 4 = 3, then there is no solution since sqrt must be positive.
396188
x^2 + 3x  = -1\nx^2 + 3x +(3/2)^2 = -1 + (3/2)^2\n(x +(3/2))^2 = (5/4)\nx + 3/2 = sqrt(5)/2 OR x + 3/2 = - sqrt(5)/2\n\n(sqrt means the positive square root) \n\nx = -(3/2) + (1/2)sqrt(5)\nx = -(3/2) - (1/2)sqrt(5)\n\n.......................................................................\n\n2x^2 +4x + 1 = 0\nx^2 + 2x = -(1/2)\nx^2 + 2x + 1 = (1/2)\n(x+1)^2 = (1/2)\n\n(x + 1) = sqrt(1/2) OR (x + 1) = -sqrt(1/2)\n\nx = -1 + sqrt(1/2) OR\nx = -1 - sqrt(1/2)\n....................................\nLW = 783\n\n\n\n(W+2)W = 783\nW^2 + 2W = 783\nW^2 + 2W  + 1 = 784\nW + 1 = sqrt(784)\n\nW = srqt(784) -1\nW + 2 = sqrt(784) + 1\n\n2(W + L) = 2(2) sqrt(784)\n                = 4* 28\n                 =   112
301868
Well, your first question is 3^(x+4)=5^(x-1).\nWhen you encounter such a problem, the first thing you should do is to apply logarithms (of base 10) to both sides of the equation! So, lg( 3^(x+4) ) = lg( 5^(x-1) ). Now, you can apply some rules involving logarithms and you'll get (lg3)(x+4)=(lg5)(x-1). Note that lg5 and lg3 are constants, and you can remove the brackets through multiplication, collect all the terms with x on one side of the equation and solve! The answer may not be "nice", but nobody said all answers are!\n\nNow, your second question is 36^(x)= 6^(x^2-3).\nEven though I did suggest using logarithms immediately for the previous question, you should take a closer look at this question to see if you can apply any "nice" results. Observe that 36=6^2, so 36^(x)= 6^(2x). So, the question becomes 6^(2x)= 6^(x^2-3). Since we have a common base of 6, we only need to analyze the powers and we have the quadratic equation (x^2)-3=2x. Can you continue from here?\n\nNOTE: For the first question, it is difficult to adopt the method of "finding a common base and equating powers" (like we did for the second question), so we have to use logarithms instead. With more experience, you should be able to know when each method can be used.
342515
4.3 x 10^5 = 430,000\n\n4.3 x 10^-5 = .000043\n\nYou just move the decimal either left or right, based off of the exponent.  If it's 5, you move it the right 5 spaces.  If it's -5, you move it to the left 5 spaces.  Fill in empty spaces with zeros.\n\nIf you have a specific question, email me, otherwise, that's really all there is to it.