Exercise 73 Page 340

For exponential equations involving exponential terms with different bases, we would first try to rewrite them to have the same base. If this is not possible, we can still solve it by using both sides as arguments of a logarithm and use the Properties of Logarithms. Consider the equation below. 2^x=3^(x+1) Let's apply the common logarithm to both sides and solve using the Properties of Logarithms.

2^x =3^(x+1)

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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)

Distr

Distribute log(3)

xlog(2) = xlog(3)+ log(3)

SubEqn

LHS-xlog(2)=RHS-xlog(2)

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

FactorOut

Factor out x

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

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

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

SubEqn

LHS-log(3)=RHS-log(3)

- log(3) = x log ( 32)

DivEqn

.LHS /log ( 32).=.RHS /log ( 32).

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

RearrangeEqn

Rearrange equation

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

Otherwise, if we have a logarithmic equation involving two logarithms of different bases, we can use the change of base formula.

If b and c are positive real numbers with b≠ 1 and c ≠ 1, then the relation shown below holds. log_c a = log_b a/log_b c In particular, log_c a = log alog c and log_c a = ln aln c.

For example, consider the equation shown below. log_4 x=log_8 4x We can rewrite both logarithms as a quotient of two logarithms having the same base. In particular, we want that the current bases can be rewritten as a power of the base we will choose. In our example equation, the bases 4 and 8 can both be rewritten as a power of base 2. Then, 2 is a good choice for the new logarithm base. log_4 x=log_8 4x ⇕ log_2 x/log_2 4=log_2( 4x)/log_2 8 Now, by rewriting 4 as 2^2 and 8 as 2^3, we can calculate the logarithms easier. Then we can solve the equation by using the Power Property of logarithms and exponentiation to undo the remaining base 2 logarithms.

log_2 x/log_2 4=log_2(4x)/log_2 8

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Solve for x

Rewrite

Rewrite 4 as 2^2

log_2 x/log_2 (2^2)=log_2(4x)/log_2 8

Rewrite

Rewrite 8 as 2^3

log_2 x/log_2 (2^2)=log_2(4x)/log_2 (2^3)

log_2(2^m)=m

log_2 x/2=log_2(4x)/3

MultEqn

LHS * 2=RHS* 2

log_2 x=2log_2(4x)/3

MultEqn

LHS * 3=RHS* 3

3log_2 x=2log_2(4x)

m* log_2(a)=log_2(a^m)

log_2 x^3=log_2((4x)^2)

CalcPow

Calculate power

log_2 x^3=log_2(16x^2)

2^(LHS)=2^(RHS)

2^(log_2 x^3)=2^(log_2(16x^2))

log_2(2^m)=m

x^3 = 16x^2

DivEqn

.LHS /x^2.=.RHS /x^2.

x=16