Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
6. Solving Exponential and Logarithmic Equations
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Exercise 4 Page 333

Choose any method that makes things easier for you. Recall that a common logarithm log has base 10, while a natural logarithm ln has base e.

rlrl a. & x = 1/4 & d. & x = sqrt(10) [1em] b. & x= - 2/3 & e. & x = e^2 [1em] c. & x = - log(3)/log ( 32) & f. & x = sqrt(27)

Practice makes perfect

When solving an equation we should choose any method that makes things easier for us. In this exercise, we will show how to solve different kinds of exponential and logarithmic equations.

a. Solving 16^x =2

Note that this equation uses 16 as a base for the exponential expression in the left-hand side, while the one on the right-hand uses 2. Since 16= 2^4, we can rewrite the equation so that both sides have the same base and we can equate the exponents. This because equal powers with same bases have same exponents. Let's give it a try.
16^x =2
â–Ľ
Solve for x
(2^4)^x =2
2^(4x) =2

a=a^1

2^(4x) =2^1
4x = 1
x = 1/4

b. Solving 2^x =4^(2x+1)

This case is similar to the previous one, but here we have two exponential expressions on each side. Since 4=2^2, we can rewrite the right-hand side and simplify so that both exponential expressions have the same base. We can then equate the exponents.
2^x =4^(2x+1)
â–Ľ
Solve for x
2^x =(2^2)^(2x+1)
2^x =2^(4x+2)
x = 4x +2
- 3x = 2
x = - 2/3

c. Solving 2^x =3^(x+1)

In this equation the bases involved do not have anything in common, as they are different prime numbers. We can use both sides as arguments of a logarithm and use the Properties of Logarithms to solve it.
2^x =3^(x+1)
â–Ľ
Solve for x

log(LHS)=log(RHS)

log(2^x) = log(3^(x+1))

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

xlog(2) = (x+1)log(3)
xlog(2) = xlog(3)+ log(3)
0= xlog(3)-xlog(2)+ log(3)
0 = x (log(3)-log(2))+ log(3)

log(m) - log(n)=log(m/n)

0 = x log ( 32)+ log(3)
- log(3) = x log ( 32)
- log(3)/log ( 32) = x
x = - log(3)/log ( 32)

d. Solving log x =1/2

In this case we can use the appropriate exponential function to undo the logarithmic function. Since the notation is just log, we know that this is a common logarithm and the base is 10.
log x = 12
â–Ľ
Solve for x

10^(LHS)=10^(RHS)

10^(log x) =10^(12)

10^(log(m))=m

x = 10^(12)

a^(1/2)=sqrt(a)

x = sqrt(10)

e. Solving ln x = 2

Once more, we can use the appropriate exponential function to undo the logarithmic function. Since the notation is ln we know that this is a natural logarithm and the base is the natural base e.
ln x = 2
â–Ľ
Solve for x

e^(LHS)=e^(RHS)

e^(ln x) = e^2

e^(ln(a))= a

x = e^2

f. Solving log_3 x = 3/2

Similarly as in the two previous cases we can use an exponential function to undo the logarithmic function. In this case the notation log_3 lets us know that we should use 3 as a base.
log_3 x = 3/2
â–Ľ
Solve for x

3^(LHS)=3^(RHS)

3^(log_3 x) = 3^(32)

3^(log_3(m))=m

x = 3^(32)

a^(m/n)=sqrt(a^m)

x = sqrt(3^3)
x = sqrt(27)