Logarithmic and exponential functions

Solve the inequality `2^x>7`, giving your answer in terms of logarithms.

`2^x>7` 

`log 2^x > log 7`

`x log 2 > log 7`

`x>frac{log7}{log2}`

Given that `lnp =2lnq-ln(3+q)` and that `q>0`, express `p` in terms of `q` not involving logarithms.

`lnp =2lnq -ln(3+q)`

`lnp =lnq^2 -ln(3+q)` 

`lnlnp =lnln(frac{q^2}{3+q})`

`p=frac{q^2}{3+q}`

Solve the inequality `3.2^(3x+2) <8`, giving your answer in terms of logarithms

`3.2^(3x+2) <8`

`3.(2^2)(2^(3x))<8` 

`8^x<2/3`

`log8^x`   

`x log 8 < log frac{2}{3}`   

`x< frac{log frac{2}{3}}{log 8}`

Use logarithms to solve the equation

`5^(x+3)=7^(x-1)`

Giving the answer correct to `3` significant figures.

`5^(x+3)=7^(x-1)`

`log(5^(x+3)) =log7^(x-1)` 

`(x+3)log5 =(x-1)log7`

`xlog5 +3log5 =xlog7 -log7`  

`x=frac{3log5 +log7}{log7 -log5} =20.1`

Solve the equation `6(4^x)-11(2^x)+4=0`, giving your answers for `x` in terms of logarithms where appropriate.

`6(4^x)-11(2^x)+4=0`

`6(2^(2x))-11(2^x)+4=0`

Let `u=2^x`

`6u^2-11u+4=0` 

`(3u-4)(2u-1)=0` 

`u=4/3` or `u=1/2`

`2^x=4/3` or `2^x=1/2`

`x=4/3` or `x=1/2`

`x=frac{2log2 -log3}{log2}` or `x=-1`

Solve the equation `ln(5x+4) =2lnx +ln6`.

`ln (5x+4) =2ln x +ln 6` 

`ln (5x+4) =ln x2 +ln 6`  

`ln (5x+4) =ln (x2+6)` 

`5x+4=x^2+6`

`x=4/3` or `x=-1/2`

But `x>0` in logarithms

So `x=4/3`

The variables `x` and `y` satisfy the equation `y=Kx^m`, where `K` and `m` are constants. The graph of `ln y` against `ln x` is a straight line passing through the points `(0, 2.0)` and `(6, 10.2)`, as shown in the diagram. Find the values of `K` and `m`, correct to `2` decimal places.

`y=Kx^m`

`ln y =ln K +ln x^m`  

`ln y =mln x +ln K` 

`Y=mX+c`

`Y=ln y`  

`m=m`

`X=ln x`  

`c=ln K`  

From the diagram:

Gradient `= m=1.37` 

`y-"intercept"=2=c`

`ln K =2`

`K=e^2=7.39`

(a) Given that `y=2^x`, show that the equation: `2^x+3(2^-x)=4` can be written form `y^2-4y+3=0`.

(b) Hence solve the equation `2^x+3(2^-x)=4` giving the values of `x` correct to `3` significant figures where appropriate.

(a)

`y=2^x`

`2^x+3(2^-x)=4` 

`2^x+3(2^x)^-1=4`

`y+3y^-1=4`

`y+3/y=4`

`y^2-4y+3=0` 

(b)

`y^2-4y+3=0` 

`y=1` or `y=3` 

`2^x=1` or `2^x=3` 

`x=0` or `x=1.58`

Given that `(1.2)^x=6^y`, use logarithms to find the value of `x/y` correct to `3` significant figures.

`(1.2)^x=6^y` 

`ln (1.2)^x =ln 6^y`

`xln 1.2 =yln 6`

`x/y=frac{ln 6}{ ln 1.2} =9.38`

The polynomial `f(x)` is defined by: `f(x)=12x^3+25x^2-4x-12`.

(a) Show that `f(-2)=0` and factorise `f(x)` completely.

(b) Given that `12(27^y)+25(9^y)-4(3^y)-12=0`, state the value of `3^y` and hence find y correct to `3` significant figures.

(a)

`f(x)=12x^3+25x^2-4x-12` 

`f(-2)=0`

`x+2` is a factor

Using long division

`f(x)=(x+2)(12x^2+x-6)=(x+2)(3x-2)(4x+3)`

(b)

`12(27^y)+25(9^y)-4(3^y)-12=0` 

`12(3^y)^3+25(3^y)^2-4(3^y)-12=0`

`f(x)=0→(x+2)(3x-2)(4x+3)=0` 

`x=-2`, `2/3`, `-4/3` 

So `3^y=-2` or `3^y=2/3` or `3^y=-4/3` but `3^y>0` 

So `3^y=2/3→y=log_3(2/3) =-0.369`

Solve the equation `|4-2^x|=10`, giving your answer correct to `3` significant figures.

`|4-2^x|=10`

`4-2^x=10` or `4-2^x=-10`

`2^x=-6` or `2^x=14` but `2^x>0`

`x=14 =3.81`

Use logarithms to solve the equation `e^x=3^(x-2)`, giving your answer correct to `3` decimal places.

`e^x=3^(x-2)`

`ln e^x =ln 3^(x-)2`

`x=(x+2)ln 3`

`x=(2ln 3)/( ln 3 -1)=22.281`

Using the substitution `u=3^x`, solve the equation `3^x+3^(2x)=3^(3x)` giving your answer correct to `3` significant figures. 

`3^x+3^(2x)=3^(3x)`

`3^x+(3x)^2=(3x)^3`

Let `u=3^x`

`u+u^2=u^3`

`u(u^2+u+1)=0`

`u=0, frac{1+sqrt5}{2}, frac{1-sqrt5}{2}` but `u=3^x>0`

`3^x=frac{1+sqrt5}{2}`

`x=frac{1+sqrt5}{2} =0.438`

The variables x and y satisfy the equation `5^y=3^(2x-4).`

(a) By taking natural logarithms, show that the graph of y against `x` is a straight line and find the exact value of the gradient of this line. 

(b) This line intersects the x-axis at P and the y-axis at `Q`. Find the exact coordinates of the midpoint of PQ.

(a)

`5^y=3^(2x-4)`

`ln 5^y =ln 3^(2x-4)`

`yln 5 =(2x-4)ln 3`

`y=(frac{2ln 3}{ln 5}) x-frac{4ln 3}{ ln 5}`

`Y=mX+c`

`Y=y`

`m=-frac{4ln 3}{ ln 5 }`

Gradient `= frac{2ln 3}{ ln 5}`

(b)

`y=frac{2ln 3}{ ln 5} x-frac{4ln 3}{ ln 5}`

When`x=0→y=-frac{4ln 3}{ ln 5}`

When `y=0→x=2`

Point P is `(2, 0)`

Point Q is `(0, -frac{4ln 3}{ ln 5})`

Midpoint `(0, -frac{2ln 3}{ ln 5})`

The variables `x` and `y`satisfy the equation `y=K(b^x)`, where `K` and `b` are constants. The graph of `ln y` against `x` is a straight line passing through the points `(2.3, 1.7)` and `(3.1, 2.1)`, as shown in the diagram. Find the values of `K` and `b`, correct to `2` decimal places. 

`y=K(b)^x`

`ln y =ln K +ln b^x`

`ln y =xln b +ln K` 

`Y=mX+c`

`Y=ln y`

`m=ln b`

`X=x`

`c=ln K`

From the diagram: 

Gradient `= m = 1/2`

`ln b =1/2`

`b=e^(1/2)=1.65`

Line passes through `(2.3,  1.7)`

`Y=mX+c`

`1.7=1/2(2.3)+c`

`c=0.55`

`ln K =0.55`

`K=e^0.55`

Variables x and y are related so that, when y is plotted on the vertical axis and x is plotted on the horizontal axis, a straight-line graph passing through the points `(2, 5)` and `(6, 11)`  is obtained. 

(a) Express `y`  in terms of `x`. 

(b) Express `y`in terms of `x`, giving your answer in the form`y=a(10^(bx))`. 

(a)

`y =mx+c`

`m=(11-5)/(6-2)=3/2`

`y =3/2x+c`

`5=3/2×2+c`

`c=2`

`y =3/2x+2`

(b)

`y =3/2x+2`

`y=10^(3/2x+2`

`y=100×10^(3/2x`

The variables x and y satisfy the equation `5^(2y)=3^(2x+1)`. By taking natural logarithms, show that the graph of `ln y` against `ln x` is a straight line, and find the exact value of the gradient of this line and state the coordinates of the point at which the line cuts the y-axis. [5]

`5^(2y)=3^(2x+1)`

`ln 5^(2y) =ln 3^(2x+1)`

`2yln 5 =(2x+1)ln 3`

`y=(frac{ln 3}{ ln 5}) x+frac{ln 3}{ 2ln 5  }`

Gradient `= frac{ln 3}{ ln 5}`

Crosses `y`-axis at `(0, frac{ln 3}{ 2ln 5})`

Solve `ln (2x+1)<= ln (x+4)` .

Solve `ln (2x+1)<= ln (x+4)`

`2x+1<= x+4`

`x≤3`

But `2x+1>0`and `x+4>0` giving `x> -1/2`

 So `-1/2 < x <= 3`

Solve `2ln (3-e^(2x))=1` , giving your answer correct to `3` significant figures. 

`2ln (3-e^(2x))=1`

`ln (3-e^(2x))=1/2`

`3-e^(2x)=e^(1/2)`

`e^(2x)=3-e^(1/2)`

`2x=ln( 3-e^(1/2))`

`x=1/2ln( 3-e^(1/2))`

Prove that the solution to the inequality `3^(2x-1)xx2^(1-3x)≥5` is `x≥frac{log (15/2)}{ log (9/8)}`

`3^(2x-1)xx2^(1-3x)≥5`

`log 3^(2x-1) +log 2^(1-3x)>= log 5`

`(2x-1)log 3 +(1-3x)log 2>= log 5`

`2xlog 3 -log 3 +log 2 -3xlog 2 ≥5`

`x≥frac{log (15/2)}{ log (9/8)}`