MODERATE EXERCISES-Logarithms

(4) Solve each equation.

  1. 100 = 50 e-x

  2. 1/4 = 52t-1

  3. ln (2x+5) = 0

  4. logx 6 = 1/3
SOLUTION:

(a) First, isolate, or solve for, the factor with the variable by dividing both sides by 50:
100 = 50 e-x ==> 2 = e-x

To solve for x, take natural logarithms (ln) of both sides, since the base of natural logarithms, e, is involved:
ln(2) = ln(e-x)

Now, by the properties of logarithms (with respect to the base "e"), ln(e-x) = -x. So we have:

ln(2) = -x, or x = -ln(2) = -0.693 (to three decimal places).

(b) In this case, we can take logs right away. We could use natural logs or common logs here. Using common logs, we get:
1/4 = 52t-1 ==> log(1/4) = log(52t-1)

To solve for t, note that we can use a property of logarithms to rewrite the right hand side:

log(52t-1) = (2t-1)log(5)

So then we have:

log(1/4) = (2t-1)log(5) or dividing by log(5), log(1/4) / log(5) = 2t-1

Solving for t, we get: 

     1      log(1/4)
t = ---  +  --------
     2       2log(5)

Using a calculator, we obtain t=0.06932 to 5 decimal places.

(c) To solve this equation, think of ln as loge and write the statement in exponential form.

So ln (2x+5) = 0 ==> loge(2x+5)=0

Writing this in exponential form gives us:
e0 = 2x+5 ==> 1 = 2x+5

Solving for x gives the answer x=-2.

(d) To solve this equation, again write it in exponential form:

logx 6 = 1/3 ==> x(1/3)= 6

Next, cube both sides to solve for x:

x(1/3)= 6 ==> [x(1/3)]3 = 63

Multiplying the exponents on the left hand side gives us x there. So after evaluating on the right hand side we get:

x = 216 as the solution.


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