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Math` `` `` ``log_4(``16^2)=log_4(256)`` ` `=log(4)(4^4) ` `=4log_4(4) ` `=4*1 ` `=4`

MathBring down the exponent `4.5lne` Remember that `lne=1 ` Simplify, `4.5*1 ` `4.5 `

MathBring down the exponent `12lne ` Remember than `lne=1 ` simplify, `12*1 ` `12 `

MathRewrite `lne^(1/2)` Bring down the exponent `(1/2)lne ` Remember that `lne=1 ` Simplify, `(1/2)*1 ` `1/2 `

MathRewrite, `lne^(3/4) ` Bring down the exponent, `(3/4)lne ` Remember that `lne=1 ` Simplify, `(3/4)*1 ` `3/4 `

MathPlug in the value of x in `g(x)` `g(x)=lne^4 ` Bring down the exponent `=4lne ` Remember that `lne=1 ` Simplify, `g(x)=4 `

MathPlug in the value for x in `g(x) ` `g(x)=lne^(5/6) ` Bring down the exponent `=(5/6)lne ` Remember that `lne=1 ` Simplify, `g(x)=5/6 `

MathPlug in the value of x in `g(x) ` `g(x)=lne^(5/2) ` Bring down the exponent `=(5/2)lne ` Remember that `lne=1 ` Simplify, `g(x)=5/2 `

MathSince there are ln's on both sides, set the terms equal to each other. `x+4=12 ` Solve for x `x=8 `

MathSince there are ln's on both sides, set the terms equal to each other. `x7=7` Solve for x `x=14`

MathSince there are ln's on both sides, set the terms equal to each other. `x^22=23` Solve for x `x^225=0 ` Factor `(x+5)(x5)=0 ` `x=5 ` and `x=5 `

MathSince there are ln's on both side, they cancel and set the two parts equal to each other. `x^2x=6` Solve for x `x^2x6=0` Factor `(x3)(x+2)=0` `x=3` and `x=2`

MathPlug in the value of x in `g(x)` `g(x)=log_a(a^2)` `=2log_a(a)` Note that `log_a(a)= ` Therefore, `g(x)=2 `

MathPlug in the value of x in `g(x)=log_b(b^3)` Bring down the exponent `=3lob_b(b)` Note that `log_b(b)=1 ` Therefore, `g(x)=3 `

MathBring down the 7 `7log_11(11) ` Note that `log_a(a)=1 ` Therefore, this simplifies down to `7 `

MathThe log of any base a of 1 is 0. (`log_a(1)=0 ` ) Therefore, this simplifies down to `0 `

MathThe log base of some value to the some value always equals 1. (`log_a(a)=1 ` ) Therefore, this simplifies down to `1 `

MathThere is a very specific rule for this, If `x=a^(log_ab) ` Then, it equals `x=b ` Apply to this problem, `9^(log_(9)15)=15 `

MathSince you have logs with the same bases on both sides, they cancel leaving you with `x+1=6 ` Solve for x `x=5 `

MathSince you have logs with the same bases on both sides, they cancel leaving you with `x3=9 ` Solve for x `x=12 `

MathSince you have logs with the same bases on both sides, they cancel leaving you with `2x+1=15 ` Solve for x `x=7 `

MathSince you have logs with the same bases on both sides, they cancel leaving you with `5x+3=12` Solve for x `x=1.8`

MathRewrite using ln rules `e^0.693=1/2 `

MathRewrite using ln rules `e^1.945=7`

MathRewrite using ln rules `e^5.521=250 `

MathRewrite using ln rules `e^0=1 `

MathRewrite into exponential form, if `e^2=7.3890 ` Then, `log_e7.3890=2 ` or `ln7.3890=2 `

MathRewrite into exponential form If, Then, `log_e1.6487=(1/2) ` or `ln1.6487=(1/2) `

MathRewrite into exponential form If, Then, `log_e0.406=0.9 ` or `ln0.406=0.9 `

MathRewrite into exponential form If Then, `log_e3=2x ` or `ln3=2x `

MathPlug in the value of x for `g(x) ` `g(x)=lne^5 ` Bring down the exponent, `=5lne ` Remember that `lne=1 ` Simplify, `g(x)=5 `

MathRewrite into exponential form If `log_4(16)=2 ` Then, `4^2=16 `

MathRewrite into exponential form If, Then, `9^2=(1/81) `

MathRewrite into exponential form If, Then, `32^(2/5)=4 `

MathRewrite in logarithmic form, If, ` ` `log_64 (8) = (1/2) ` Then, `64^(1/2)=8 `

MathRewrite in exponential form If, Then, `log_5(125)=3 `

MathRewrite in exponential form If, Then, `log_(9)27=3/2 `

MathRewrite in exponential form If, Then, `log_(4)(1/64)=3 `

MathRewrite in logarithmic form, If, Then, `log_24(1)=0 `

MathPlug in the value of x in `f(x) ` `f(x)=log_2(64) ` `y=log_2(64) ` `2^y=64 ` `2^y=2^6 ` `y=6 `

MathPlug in the value of x in `f(x) ` `f(x)=log_25(5) ` `y=log_25(5) ` `25^y=5 ` `5^(2y)=5^1 ` `2y=1 ` `y=1/2 `

MathPlug in the value of x in `f(x) ` `f(x)=log_8(1) ` Remember that for all `log_a(1)=0 ` simplify, `f(x)=0 `

MathPlug in the value of x for `f(x) ` `f(x)=log10 ` Assume a base of 10 `f(x)=log_(10)10 ` Remember that `log_a(a)=1 ` Simplify, `f(x)=1 `

MathYou need to find the absolute extrema of the function, hence, you need to differentiate the function with respect to x, such that: `f'(x) = (2x^3  6x)'` `f'(x) = 6x^2  6` You need to solve for x...

MathYou need to find out the absolute extrema of the given function, hence, you need to differentiate the function with respect to x, such that: `y' = 3*(2/3)*x^(2/3  1)  2` You need to solve for x...

MathYou need to find out the absolute extrema of the given function, hence, you need to differentiate the function with respect to x, such that: `g'(x) = (root(3)(x))'` `g'(x) = (1/3)x^(1/3  1) g'(x)...

MathYou need to evaluate the absolute extrema of the function, hence, you need to differentiate the function with respect to t, using the quotient rule, such that: `g'(t) = ((t^2)'*(t^2 + 3) ...

MathYou need to evaluate the absolute extrema of the function, hence, you need to differentiate the function with respect to x, using the quotient rule, such that: `f'(x) = ((2x)'(x^2 + 1)  2x*(x^2 +...

MathYou need to find the derivative of the function, using the quaotient rule, such that: `h'(s) = (1'*(s  2)  1*(s 2)')/((s2)^2)` `h'(s) = (0*(s  2)  1*1)/((s2)^2)` `h'(s) =1/((s2)^2)` You...

MathYou need to find the derivative of the function h(t), using the quotient rule, such that: `h'(t) = (t'*(t + 3)  t*(t +3)')/((t + 3)^2)` `h'(t) = ((t + 3)  t*1)/((t + 3)^2)` `h'(t) = (t + 3 ...