Answer:
To solve an exponential equation, take the log of both sides, and solve for the variable.
Step-by-step explanation:
Example 1: Solve for x in the equation tex2html_wrap_inline119 .
Solution:
Step 1: Take the natural log of both sides:
displaymath121
Step 2: Simplify the left side of the above equation using Logarithmic Rule 3:
displaymath123
Step 3: Simplify the left side of the above equation: Since Ln(e)=1, the equation reads
displaymath127
Ln(80) is the exact answer and x=4.38202663467 is an approximate answer because we have rounded the value of Ln(80)..
Check: Check your answer in the original equation.
displaymath131
Answer:
Step-by-step explanation:
Let b be the number of balloons and h be number of party hats.
We have been given that balloons cost $0.50 each, so cost of b balloons will be 0.50b.
We are also told that party hats each cost $1.25, so cost of h party hats will be 1.25h.
Further, Joe wants to spend exactly $20 on the party supplies. We can represent this information in an equation as:
Therefore, the equation represents the number of balloons ( b) and the number of party hats ( h) that Joe can buy spending exactly $20.
Answer:
x=4/3
Step-by-step explanation:
Answer:
The answer is B
Step-by-step explanation:
3 to the forth power = 81
3 to the second power = 9
81/9 = 9
1. We use the recursive formula to make the table of values:
f(1) = 35
f(2) = f(1) + f(2-1) = f(1) + f(1) = 35 + 35 = 70
f(3) = f(1) + f(3-1) = f(1) + f(2) = 35 + 70 = 105
f(4) = f(1) + f(4-1) = f(1) + f(3) = 35 + 105 = 140
f(5) = f(1) + f(5-1) = f(1) + f(4) = 35 + 140 = 175
2. We observe that the pattern is that for each increase of n by 1, the value of f(n) increases by 35. The explicit equation would be that f(n) = 35n. This fits with the description that Bill saves up $35 each week, thus meaning that he adds $35 to the previous week's value.
3. Therefore, the value of f(40) = 35*40 = 1400. This is easier than having to calculate each value from f(1) up to f(39) individually. The answer of 1400 means that Bill will have saved up $1400 after 40 weeks.
4. For the sequence of 5, 6, 8, 11, 15, 20, 26, 33, 41...
The first-order differences between each pair of terms is: 1, 2, 3, 4, 5, 6, 7, 8...since these differences form a linear equation, this sequence can be expressed as a quadratic equation. Since quadratics are functions (they do not have repeating values of the x-coordinate), therefore, this sequence can also be considered a function.
