Solving Exponential Equations Using Logarithms. Can someone help! Will give 5 stars!?
Logarithms are always confusing at first. Learning them is a matter of practice. Understanding will come. Let’s walk slowly through these problems. I am going to use “log” to mean “logarithm to the base b.” ----------------------------- 2^x = 30 Whenever the variable you want to isolate is in an exponent, it is necessary to take a log. log(2^x) = log(30) Remember now that a logarithm is an exponent. If b and q are positive numbers, then b^[something] = q. By definition, [something] is log q. Remember also that b^(a^c) = (b^a)^c = b^(ac). Since a logarithm is an exponent, log(q^a) = a * log(q). Knowing this, we can modify log(2^x) = log(30) to make it x * log(2) = log(30) Then we just do some ordinary algebra. x = log(30) / log(2) ≈ 4.91 ----------------------------- 5^(x - 1) = 3^x Take logs. log[5^(x – 1)] = log(3^x) (x – 1) log(5) = x * log(3) x * log(5) – log(5) = x * log(3) x * log(5) – x * log(3) = log(5) x[log(5) – log(3)] = log(5) x = log(5) / [log(5) – log(3)] We may decide to consider ourselves done at this point. Alternatively, we can remember that subtracting logarithms is equivalent to dividing. Therefore we may, if we wish, finish with x = log(5) / [log(5/3)] Either way, we can get out our calculator and compute x ≈ 3.15 ----------------------------- 3.53^(x + 1) = 65.4 Take logs, and make the exponent into a multiplier. (x + 1) log(3.53) = log(65.4) x * log(3.53) = log(65.4) - log(3.53) = log(65.4 / 3.53) = log(18.5269121813031) In this case, the base of the logarithm matters. I will not finish the calculation. I leave it to you, if you wish, to pick a base you like and finish the calculation. ----------------------------- 16^(x - 4) = 33 - x I know of no way to solve for x when it appears as both an exponent and a term. ----------------------------- 7^(x - 2) = 5^x (x - 2) log(7) = x * log(5) x * log(7) - x * log(5) = 2 log(7) x [log(7) - log(5)] = 2 log(7) x [log(7) - log(5)] = 2 log(7) x = 2 log(7) / [log(7) - log(5)] = log(49) / log(7/5) ≈ 11.57
Help with logarithms ?
To convert logarithms to have common base, consider the following: log [base x] (y) = log [base z] (y) / log [base z] (x) so for example: log[2](4) = log4 / log2 = 2 Also remember: 2 * logx = log(x²) 1) log(base 3) x = 9 log(base x) 3 logx / log3 = 9 (log3 / logx) logx * logx = (3 * 3) * log3 * log3 (logx)² = (3log3)² (logx)² = (log(3³))² logx = log(3³) x = 3³ = 27 2) log(base 2) x = log(base 4) (x+6) logx / log2 = log(x+6) / log4 logx * log4 / log 2 = log(x+6) logx * 2 = log(x+6) log(x²) = log(x+6) x² = x + 6 x² - x - 6 = 0 (x - 3)(x + 2) = 0 x = 3 or -2 remember there was log[2](x), and you can't log a negative number, so x cannot equal -2 thus: x = 3 3) 3^(x+1) * 2^(x-2) = 21 log [ 3^(x+1) * 2^(x-2) ] = log21 log[3^(x+1)] + log[2^(x-2)] = log21 (x+1)log[3] + (x-2)log[2] = log21 xlog3 + log3 + xlog2 - log2 = log21 x(log3 + log2) = log21 - log3 + log2 x = (log21 - log3 + log2) / (log3 + log2) x = log(21*2/3) / log(3*2) x = log(14) / log(6) ... feel free to calculate that if you want 4) 4(3^2x) = e^x ln[4(3^2x)] = ln[e^x] << note I'm taking the natural logarithm this time ln[4] + ln[(3^2x)] = x log[e] ln[4] + 2x ln[3] = x ln[4] = x - 2x ln[3] ln4 = x(1 - 2ln3) x = ln4 / (1 - 2ln3) ... feel free to calculate that if you want For 5&6, bear in mind that: e^2x = e^(x+x) = e^x * e^x = (e^x)² 5) 2e^2x - 3e^x = 2 2y² - 3y = 2 2y² - 3y - 2 = 0 (2y + 1)(y - 2) = 0 y = 2 or -½ y = e^x e^x = 2 or -½ ln[e^x] = ln2 or ln(-½) x = ln2 or ln(-½) can't take a logarithm of a negative number thus x = ln2 6) e^3x + 2e^x = 3e^2x y³ + 2y = 3y² y³ - 3y² + 2y = 0 y(y² - 3y + 2) = 0 y(y-2)(y-1) = 0 y = 0 or 1 or 2 y = e^x e^x = 0 or 1 or 2 x = ln0 or ln1 or ln2 ln0 doesn't exist, and ln1 = 0, thus: x = 0 or ln2
Im having some trouble with math 12 logarithm unit?
log[2](15/2) => log[2](3 * 5 / 2) => log[2](3) + log[2](5) - log[2](2) => y + x - 1 9 * log[27](x) - 4 * log[9](x) => 9 * log(x) / log(27) - 4 * log(x) / log(9) => 9 * log(x) / (3 * log(3)) - 4 * log(x) / (2 * log(3)) => 3 * log(x) / log(3) - 2 * log(x) / log(3) => log(x) / log(3) => log[3](x)
Logarithm and exponent questions? 10 points WILL be rewarded :)?
log(base3) (x+3)^2 - log(base3)(x+1)= log (base 3) 2^3 log(base3) (x+3)^2/(x+1)= log (base 3) 2^3 then (x+3)^2/(x+1)=8 x^2+6x+9=8x+8 x^2-2x+1=0 x=1 double root 3^(2x) 5 ^x= 7^(3x+1) (45)^x=7*(343)^x then (45/343)^x=7 x=(log(base10) 7)/(log(base 10) (45/343)) 5*3^x-40*3^(-x)=17 substitute 3^x=y 5y-40/y=17 but y is not zero ( because 3^x do not be 0) 5y^2-17y-40=0 (5y+8)(y-5)=0 then y=5 or y=-8/5 is not accepted only y=5 is good then 3^x=5 then x=log5/log 3
I want the answer to this log question, with working please?
I've used square brackets to indicate the base and I've used the following rules : log[A]B = log[C]B/log[C]A, where C is any base. log[C](A^N) = Nlog[C]A, where C is any base. A^(log[A]B) = B (A^B)^N = A^(BN) It also seems apparent that if everything is put to the base 5, then the third rule will "rule OK". Now I'll take each part in turn, one step at a time. Take it slowly so you understand the rules used. (1) 2/log[3](25) = 2/log[3](5^2) = 2/(2log[3]5) = 1/log[3]5 = 1/(log5/log3) = log3/log5 = log[5]3 Now, 625^(log[5]3) = (5^4)^(log[5]3) = 5^(4log[5]3) = 5^{log[5](3^4)} = 3^4 = 81. (2) log[125](27) = log(27)/log(125) = log(3^3)/log(5^3) = 3log3/(3log5) = log3/log5 = log[5]3 Now, 25^(log[5]3) = (5^2)^(log[5]3) = 5^(2log[5]3) = 5^{log[5](3^2)} = 3^2 = 9. (3) Because of the integers we already have, I can't help but feel that this last one is incorrectly given. My intuition says that it should be 5^(3/(log base 6) 125), where a 3 should be where the second 5 is. Is that correct? If not, then the rest of this is wasted. 3/log[6](125) = 3/{log(125)/log6} = 3/{log(5^3)/log6} = 3/(3log5/log6) = log6/log5 = log[5]6 Now, 5^(log[5]6) = 6. Thus, P = 81 + 9 + 6 = 96.
Which one is bigger? 3^pi or pi^3? How can I do mental math with this?
You have two numbers [math]x[/math] and [math]y[/math] with [math]x>y[/math] and the question is, which is larger [math]x^y[/math] or [math]y^x[/math]. To answer this, consider instead the function [math]f(x) = x^{\frac{1}{x}}[/math] for [math]x > 0[/math]. The derivative of this function is [math]f'(x) = x^{\frac{1}{x} - 2 } \left( 1 - \log x \right)[/math]This is positive for [math]\log x < 1 \implies x < e[/math] and negative for [math]\log x > 1 \implies x > e[/math]. This implies that this function has a maximum at [math]x = e[/math]. Further, we note that for very large [math]x[/math], the function limits to [math]1[/math]. Then, this function is less than one, only if [math]x < 1[/math]. Great! We have all the information we need to answer your question, and much much more. Let us take cases. Let [math]x > y > e[/math]. Then, by the properties of the function, [math]f(y)>f(x)[/math]. This implies [math]x^{\frac{1}{x}} < y^{\frac{1}{y}} \implies x^y < y^x[/math]. This is true for any two numbers greater than [math]e[/math]. In particular, the two numbers you have at hand are [math]\pi[/math] and [math]3[/math] both of which are greater than [math]e \approx 2.71[/math]. Further, [math]\pi > 3[/math]. Hence, [math]\pi^3 < 3^\pi[/math]. Let [math]e>x>y[/math]. Then, by the properties of the function, [math]f(x)>f(y)\implies x^y > y^x[/math]. For instance, since [math]e>0.75>0.5[/math], we must have [math]0.75^{0.5} > 0.5^{0.75}[/math] which is true (as you can check).When [math]x>e>y[/math], then the question is trickier and we cannot answer which is larger in general. However, at least in one case, we can answer this. If we have [math]x>e>1>y[/math], then we have [math]f(y) < f(x)\implies y^{1/y} >x^{1/x}\implies x^y > y^x[/math].
How to solve these log functions?
log(x) + log(x - 15) = 2 log(x * (x - 15)) = 2 x * (x - 15) = 10^2 x^2 - 15x = 100 x^2 - 15x - 100 = 0 x = (15 +/- sqrt(225 + 400)) / 2 x = (15 +/- sqrt(625)) / 2 x = (15 +/- 25) / 2 x = 40/2 , -10/2 x = 20 , -5 x = 20 is the only solution that works in the original equation log[5](m) = 1 + log[5](m + 2) log[5](m) - log[5](m + 2) = 1 log[5](m / (m + 2)) = 1 m / (m + 2) = 5^1 m / (m + 2) = 5 m = 5m + 10 -10 = 5m - m -10 = 4m m = -5/2 log(x + 9) = log(4x) x + 9 = 4x 9 = 3x 3 = x -3 * log[3](-9x) = -3 log[3](-9x) = 1 -9x = 3^1 -9x = 3 x = -3/9 x = -1/3