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  • Math
    To solve the equation `log_3(x)=log_9(6x)`, we may apply logarithm properties. Apply the logarithm property: `log_a(b)= (log_c(b))/log_c(a)` on `log_3(x)` , we get:...

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  • Math
    Solve `log_2(x+1)=log_8(3x) ` : Rewrite using the change of base formula: `(ln(x+1))/(ln(2))=(ln(3x))/(ln(8)) ` `(ln(8))/(ln(2))=(ln(3x))/(ln(x+1)) ` But `ln(8)=ln(2^3)=3ln(2) ` so:...

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  • Math
    To solve the equation: `10^(3x-8)=2^(5-x)` , we may take "ln" on both sides. `ln(10^(3x-8))=ln(2^(5-x))` Apply natural logarithm property: `ln(x^n) = n*ln(x)` . `(3x-8)ln(10)=(5-x)ln(2)` Let...

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  • Math
    `3^(x+4) = 6^(2x-5)` To solve, take the natural logarithm of both sides. `ln (3^(x+4)) = ln (6^(2x-5))` To simplify each side, apply the logarithm rule `ln (a^m) =m*ln(a)` . `(x+4)ln(3) = (2x-5) ln...

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  • Math
    To evaluate the given equation `log_6(3x)+log_6(x-1)=3` , we may apply the logarithm property: `log_b(x)+log_b(y)=log_b(x*y)` . `log_6(3x)+log_6(x-1)=3` `log_6(3x*(x-1))=3` `log_6(3x^2-3x)=3` To...

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  • Math
    We are asked to solve the following equation: `4ln(-x)+3=21` Use basic algebraic rules to isolate the term with the logarithm: `4ln(-x)=18` `ln(-x)=9/2` Exponentiate both sides with base e:...

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  • Math
    To evaluate the given equation `log_4(-x)+log_4(x+10)=2` , we may apply the logarithm property: `log_b(x)+log_b(y)=log_b(x*y)` . `log_4(-x)+log_4(x+10)=2` `log_4((-x)*(x+10))=2`...

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  • Math
    To evaluate the given equation `log_2(x-4)=6` , we may apply the logarithm property: `a^(log_a(x))=x` . Raised both sides by base of `2` . `2^(log_2(x-4))=2^6` `x-4=64` Add `4` on both sides to...

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  • Math
    To evaluate the given equation `1/3log_5(12x)=2` , we may apply logarithm property: `n* log_b(x) = log_b(x^n)` . `log_5((12x)^(1/3))=2` Take the "log" on both sides to be able to apply the...

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  • Math
    We are asked to solve `log_8(5-12x)=log_8(6x-1)` Exponentiating both sides with base 8 we get: 5-12x=6x-118x=6 x=1/3. This is in the domain of both expressions of the equality (the domain for the...

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  • Math
    To evaluate the equation `log_6(3x-10)=log_6(14-5x)` , we apply logarithm property: `a^(log_a(x))=x` . Raised both sides by base of `6` . `6^(log_6(3x-10))=6^(log_6(14-5x))` `3x-10=14-5x` Add `10`...

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  • Math
    To solve the equation `log_3(18x+7)=log_3(3x+38)` , we apply logarithm property: `a^(log_a(x))=x` . Raised both sides by base of `3` . `3^(log_3(18x+7))=3^(log_3(3x+38))` `18x+7=3x+38` Subtract 7...

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  • Math
    `log(12x-11)=log(3x+13)` Using the property of logarithmic equality, `12x-11=3x+13` `=>12x-3x=13+11` `=>9x=24` `=>x=24/9` `=>x=8/3` Let's plug back the solution in the equation to check...

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  • Math
    `log_5(2x-7)=log_5(3x-9)` Using one to one property of logarithms, `2x-7=3x-9` `=>2x-3x=-9+7` `=>-x=-2` `=>x=2` Let's plug back the solution in the equation, `log_5(2*2-7)=log_5(3*2-9)`...

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  • Math
    To evaluate the equation `ln(x+19)=ln(7x-8)` , we apply natural logarithm property: `e^(ln(x))=x` . Raise both sides by base of `e` . `e^(ln(x+19))=e^(ln(7x-8))` `x+19=7x-8` Subtract `7x` from both...

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  • Math
    `ln(4x-7)=ln(x+11)` Using one to one property of logarithms, `4x-7=x+11` `=>4x-x=11+7` `=>3x=18` `=>x=18/3` `=>x=6` Plug back the solution in the equation to check the solution,...

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  • Math
    To solve the equation `log_5(5x+9)=log_5(6x)` , we apply logarithm property: `a^(log_a(x))=x` . Raise both sides by base of `5` . `5^( log_5(5x+9))=5^(log_5(6x))` `5x+9=6x` Subtract `5x` from...

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  • Math
    For the given equation `2^(0.1x)-5=12` , we may simplify by combining like terms. Add `5` on both sides of the equation. `2^(0.1x)-5+5=12+5` `2^(0.1x)=17` Take the "`ln` " on both sides to be able...

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  • Math
    For the given equation `0.5^x-0.25=4` , we may simplify by combining like terms. Add `0.25` on both sides of the equation. `0.5^x-0.25+0. 25=4+0.25` `0.5^x=4.25` Take the "`ln` " on both sides to...

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  • Math
    For the given equation `10^(3x)+4 =9` , we may simplify by combining like terms. Subtract 4 from both sides of the equation. `10^(3x)+4-4 =9-4` `10^(3x)=5` Take the "ln" on both sides to be able to...

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  • Math
    To solve the given equation `7^(6x)=12` , we may take "`ln` " on both sides of the equation. `ln(7^(6x))=ln(12)` Apply natural logarithm property: `n*ln (x)=ln (x^n)` . `6x*ln(7)=ln(12)` Divide...

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  • Math
    To solve the given equation `7^(3x)=18` , we may take "ln" on both sides of the equation. `ln(7^(3x))=ln(18)` Apply natural logarithm property:` ln (x^n) = n*ln (x)` . `3xln(7)=ln(18)` Divide both...

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  • Math
    To solve the given equation `8^x=20` , we may take "ln" on both sides of the equation. `ln(8^x)=ln(20)` Apply natural logarithm property: `ln (x^n) = n*ln (x)` . `xln(8)=ln(20)` Divide both sides...

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  • Math
    To evaluate the given equation `25^(10x+8)=(1/125)^(4-2x)` , we may apply `25=5^2` and `1/125=5^(-3)` . The equation becomes: `(5^2)^(10x+8)=(5^(-3))^(4-2x)` Apply Law of Exponents: `(x^n)^m =...

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  • Math
    To evaluate the given equation `10^(3x-10)=(1/100)^(6x-1)` , we may apply `100=10^2` . The equation becomes: `10^(3x-10)=(1/10^2)^(6x-1)` Apply Law of Exponents: `1/x^n = x^(-n)` ....

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  • Math
    To evaluate the given equation `36^(5x+2)=(1/6)^(11-x)` , we may apply `36=6^2` and `1/6=6^(-1)` . The equation becomes: `(6^2)^(5x+2)=(6^(-1))^(11-x)` Apply Law of Exponents: `(x^n)^m = x^(n*m)`...

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  • Math
    `3^(3x-7)=81^(12-3x)` To solve, factor 81. `3^(3x-7)=(3^4)^(12-3x)` To simplify the right side, apply the exponent rule `(a^m)^n=a^(m*n)` . `3^(3x-7)=3^(4*(12-3x))` `3^(3x-7)= 3^(48-12x)` Since...

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  • Math
    To evaluate the given equation `4^(2x-5)=64^(3x)` , we may let `64 =4^3` . The equation becomes: `4^(2x-5)=(4^3)^(3x)` . Apply Law of exponents: `(x^n)^m = x^(n*m)` . `4^(2x-5)=4^(3*3x)`...

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  • Math
    `27^(4x-1)=9^(3x+8)` To solve, factor the 9 and 27. `(3^3)^(4x-1)=(3^2)^(3x+8)` To simplify each side, apply the exponent rule `(a^m)^n=a^(m*n)` . `3^(3*(4x-1))=3^(2*(3x+8))` `3^(12x-3)=3^(6x+16)`...

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  • Math
    `8^(x-1)=32^(3x-2)` To solve, factor 8 and 32. `(2^3)^(x-1)=(2^5)^(3x-2)` To simplify each side, apply the exponent rule `(a^m)^n = a^(m*n)` . `2^(3*(x-1)) = 2^(5*(3x-2))` `2^(3x-3) = 2^(15x-10)`...

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  • Math
    `7^(3x+4)=49^(2x+1)` To solve, factor the 49. `7^(3x+4)=(7^2)^(2x+1)` To simplify the right side, apply the exponent property `(a^m)^n=a^(m*n)` . `7^(3x+4)=7^(4x+2)` Since the two sides have the...

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  • Math
    `5^(x-4)=25^(x-6)` To solve, factor the 25. `5^(x-4)=(5^2)^(x-6)` To simplify the right side, apply the exponent rule `(a^m)^n = a^(m*n)` . `5^(x-4)=5^(2*(x-6))` `5^(x-4)=5^(2x-12)` Since both...

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  • Math
    To determine the power function `y=ax^b` from the given coordinates: `(5,10)` and `(12,81)` , we set-up system of equations by plug-in the values of x and y on `y=ax^b.` Using the coordinate...

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  • Math
    To determine the power function `y=ax^b` from the given coordinates: `(4,8) ` and `(8,30)` , we set-up system of equations by plug-in the values of `x` and `y` on `y=ax^b` . Using the coordinate...

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  • Math
    We are asked to write a power function whose graph includes the points (3,14) and (9,44): `14=a3^b,44=a9^b` From the first equation we get: `a=14/(3^b)` Then `44=(14/(3^b))*9^b` `44=14*3^b`...

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  • Math
    We are asked to write a power function whose graph includes the points (2,3) and (6,12). Substitute the given x,y values into the base equation to get two equations in the two unknowns a,b. Solve...

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  • Math
    We are asked to write a power function whose graph includes the points (5,9) and (8,34). Substitute the given x,y pairs into the base model to get two equations with the two unknowns a,b. Solve the...

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  • Math
    We are asked to write the equation for a power function whose graph passes through the points (4,3) and (8,15). We substitute the known values of x and y into the basic equation to get two...

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  • Math
    The given two points of the exponential function are (1,40) and (3,640). To determine the exponential function `y=ab^x` plug-in the given x and y values. For the first point (1,40), plug-in x=1...

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  • Math
    The given two points of the exponential function are (1,2) and (3,50). To determine the exponential function `y=ab^x` plug-in the given x and y values. For the first point (1,2), plug-in x=1 and...

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  • Math
    The given two points of the exponential function are (3,27) and (5,243). To determine the exponential function `y=ab^x` plug-in the given x and y values. For the first point (3,27), the values of x...

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  • Math
    To determine the power function `y=ax^b` from the given coordinates: `(3,1)` and `(5,4)` , we set-up system of equations by plug-in the values of x and y on `y=ax^b` . Using the coordinate `(3,1)`...

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  • Math
    The given two points of the exponential function are (2,24) and (3,144). To determine the exponential function `y=ab^x` plug-in the given x and y values. For the first point (2,24), the values of x...

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  • Math
    The given two points of the exponential function are (1,3) and (2,12). To determine the exponential function `y=ab^x` plug-in the given x and y values. For the first point (1,3), plug-in x = 1 and...

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  • Math
    We are asked to write the equation of the parabola with directrix x=-1/18 and vertex at the origin: The equation for a parabola with vertex at the origin and focus (a,0) is `y^2=4ax ` Since the...

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  • Math
    We are asked to write the equation of the parabola with directrix y=5/12 and vertex at the origin: The equation for a parabola with vertex at the origin and focus (0,a) is `x^2=4ay ` Since the...

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  • Math
    We are asked to write the equation of the parabola with vertex at the origin and directrix x=11: The equation of a parabola with vertex at the origin and focus at (a,0) is `y^2=4ax ` ; the parabola...

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  • Math
    A parabola with directrix at` x=a` implies that the parabola may opens up sideways towards to the left or right. The position of the directrix with respect to the vertex point can be used to...

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  • Math
    A parabola with directrix at `y=k` implies that the parabola may opens up towards upward or downward direction. The position of the directrix with respect to the vertex point can be used to...

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  • Math
    A parabola with directrix at `y=k` implies that the parabola may open up towards upward or downward direction. The position of the directrix with respect to the vertex point can be used to...

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