Exercise 98 Page 340

Practice makes perfect

We want to solve the following equation.

2(x-1)^2=18

We can see that it is quadratic. Note that since both sides of the equation are non-negative, we can take the square root of both sides. Before we do that, let's divide both sides of the equation by 2 to isolate the power on the left-hand side of the equation. 2(x-1)^2=18 ⇔ (x-1)^2=9 Now, we will take the square roots and solve. Keep in mind that we need to consider the positive and negative solutions.

(x-1)^2 = 9

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((x-1)^2) = sqrt(9)

SqrtToAbs

sqrt(a^2)=|a|

|x-1| = sqrt(9)

CalcRoot

Calculate root

|x-1| = 3

lc x-1 ≥ 0:x-1 = 3 & (I) x-1 < 0:x-1 = - 3 & (II)

lcx-1=3 & (I) x-1=-3 & (II)

(I), (II): LHS+1=RHS+1

lx_1=4 x_2=-2

Therefore, the solutions for the equation are 4 and -2.

We are given an exponential equation. When bases are not the same, we can solve it by taking the logarithm of each side of the equation.

m=n ⇔ log m = log n Note that in order to take their logarithms, both m and n must be positive numbers. In our case, let's first substitute 3 from both sides of the equation to isolate the exponential expression. 2^x+3=10 ⇔ 2^x=7 Now, let's take the logarithm of each side.

2^x=7

log(LHS)=log(RHS)

log 2^x= log 7

We obtained a logarithm of a power. To isolate x we can use the Power Property of Logarithms.

log 2^x= log 7

log(a^m)= m*log(a)

xlog 2= log 7

DivEqn

.LHS /log 2.=.RHS /log 2.

x=log 7/log 2

UseCalc

Use a calculator