Need help on logarithms?
you people are a great help! please help me on these too: 1) if f(x)=log3^x, what are the domain and range of x? 2) solve for x: log x + log (x+3)=1 3) if logx-2^125=3, find x. 4) solve for x: log (x - 2) + log (2x-3)= 2 log x 5) evaluate: logb^1
Write the ratio 4 to 12 in two other ways?
4:12 4/12 :)
2^(x+1) = 3^(x-1) ---> logarithms!!!?
Q1 2^(x +1) = 3^(x - 1) take the log of both sides: log[ 2^(x + 1) ] = log [ 3^(x - 1) ] now use the property of logs => log(a^b) = b * log(a) (x + 1) * log(2) = (x - 1) * log(3) expand the (x + 1) and (x - 1) terms using distributive property => a*(b + c) = a*b + a*c x * log(2) + 1 * log(2) = x * log(3) - 1 * log(3) move all x terms to one side and all non-x terms to the other: x * log(2) - x * log(3) = -1 * log(3) - 1 * log(2) common factor the x: x * [ log(2) - log(3) ] = -log(3) - log(2) divide both sides by [ log(2) - log(3) ]: x = [ -log(3) - log(2) ] / [ log(2) - log(3) ] = 4.41902 Q2 [ 4^x ] * [ 3^(2x + 1) ] = 6^(x + 1) as in Q1, take logs of both sides: log [ (4^x)(3^(2x + 1) ) ] = log [ 6^(x + 1) ] use the log property of products on the left side => i.e log [ a * b] = log(a) + log(b) log[ 4^x ] + log[ 3^(2x + 1) ] = log[ 6^(x + 1) ] now use this log property on both sides => log[ a^b ] = b * log[ a ] x * log [ 4 ] + (2x + 1) * log[ 3 ] = (x + 1) * log[ 6 ] expand the (2x + 1) and (x + 1) terms: x * log [ 4 ] + 2x * log[ 3 ] + log[ 3 ] = x * log [ 6 ] + log [ 6 ] collect all x terms on one side and move all non-x terms to the other: x * log [ 4 ] + 2x * log [ 3 ] - x * log [ 6 ] = log [ 6 ] - log [ 3 ] common factor the x: x * ( log [ 4 ] + 2 * log [ 3] - log [ 6 ] ) = log [ 6 ] - log [ 3] divide both sides by ( log[ 4 ] + 2 * log [ 3 ] - log [ 6 ] ): x = ( log [ 6] - log [ 3 ] ) / ( log [ 4] + 2 * log [ 3] - log [ 6 ] ) = 0.3868
SOLVE FOR X (LOGARITHMIC PROBLEM)?
1) Since both has same base a, we can write this as log x = 3 log 2 which makes log x = log 2^3 or log x = log 8 now raise both sides by 10 10^log x = 10^log 8 since 10^log = 1, the logs drop out and we have x = 8 I won't show all these steps with the others. 2) log(x)(x-3) = log(x^2-12) x^2-3x = x^2-12 -3x = -12 x = 4 3) log(x-2)(5) / (x-2) = log x 5x-10 / x-2 = x 5x-10 = x^2-2x x^2-7x+10 = 0 (x-2)(x-5) = 0 x=2 x=5 4) log(2)(x) = log(4^3) / 3 2x = 48/3 x = 8 5) Oops! Didn't see that some of the problem is missing. chow! :)
If log x/log y = 3/2 and log(xy) = 5. find the value of x and y?
logx/logy = 3/2, log(xy) = 5,logx = 3/2 logy —————->eqn 1log(xy) = log x + log ylog x + log y = 5, —————->eqn 2Substitute eqn 2 in eqn 1!3/2 logy + log y = 5,5/2 log y = 5 , cancel 5 on both sideslog y = 2log has base 10 so y is given byy = 10 pow(2)y = 100Substitute y = 100 in eqn 2log x + logy = 5log x + log (100) = 5log x + 2 = 5log x = 3 , log has a base of 10 x given asx = 10 pow(3)x = 1000.Hence x=1000, y = 100
Exponentials and Logarithms Questions?
a^x = 10^(2x+1) x log(a) = 2x + 1 x = 1 / (log(a) - 2) Answer (1a): x = 1 / (log(a) - 2) 2log(2x) = 1 + log(a) log(4x²) = log(10a) x = √(5a/2) ............... discard the negative root Answer (1b): x = √(5a/2) 2^x = 3^y x + y = 1 2^x = 3^(1-x) 2^x = 3/3^x 6^x = 3 x = log(3)/log(6) Answer (2): see above log2(x+2) - log2(x) = 3 log2((x+2)/x) = 3 (x+2)/x = 8 x = 2/7 Answer (3): x = 2/7
If log 2 =a, log 3=b, log 7=c and 6^x=7^x+4, then what is the value of x?
‘If log 2 =a, log 3=b, log 7=c and 6^x=7^x+4, then what is the value of x?’ The question is badly written.Do you want [math]x[/math] in terms of [math]a[/math], [math]b[/math] and [math]c[/math]?[math]6^x = 7^{x + 4}[/math]Take logarithms to whatever (single) base the logarithms [math]a[/math], [math]b[/math] and [math]c[/math] are given, assuming the logarithms are defined.[math]x\log6 = (x + 4)\log7[/math][math]\implies x = \cfrac{4\log7}{\log6 - \log7}[/math][math]\implies x = \cfrac{4\log7}{\log(2\times3) - \log7}[/math][math]\implies x = \cfrac{4\log7}{\log2 + \log3 - \log7}[/math][math]\implies x = \cfrac{4c}{a + b - c}[/math]You haven’t explicitly mentioned that you want the answer in terms of these numbers. Interestingly, the value of [math]x[/math] is independent of the base of the logarithms (again, assuming the logarithms are defined).[math]x = -50.493\,716[/math]This a direct consequence of the fact that the original equation can be solved without knowing the logarithm values.
If x = 1 + log 2 - log 5, y = 2 log 3 and z = log a - log 5; what is the value of a, if x + y=2z?
Just simply put the values of x, y, z in given equation and apply basic mathematics=> (1 + log 2- log 5)+(2 log 3)=2 log a-2 log 5=> 1+log 2+log 5+ 2 log 3=2 log aHere we can write 1=log 10 as considering log natural base 10.and use a log b =log b^a=> log 10 +log 2+ log 5 + log 9= 2 log a{ using log a+ log b= log (a+b) }=> 2 log a=log(10*2*5*9)=> 2 log a= log (10*10*3*3)=> a^2=10*10*3*3=> a=10*3=>a=30
If log2 = a and log 3 = b , then express log 25/8 in terms of a and b?
What is the sum of first ten terms of the series log 2 + log 4 + log 8 +?
log 2 + log 4 + log 8 + … + log 1024 (10th term would be 2^10)= log 2 + log (2^2) + log (2^3) + … + log (2^10)= log 2 + 2 log 2 + 3 log 2 + …. + 10 log 2= 55 log 2Hope it helped :)