This leads to the solution of an exponential equation, as it is required to find the exponent of the power to which 3 must be raised to equal. Hence

3x

27.

This equation may be solved in the same way as the preceding problems, but whenever the numbers on each side of the equation can be expressed as powers of the same number, the equation can be solved by inspection without the use of logarithms.

157. Let P = the principal at compound interest r = the rate per cent

n = the number of years

A the amount at the end of n years.

=

Then at the end of the first year, A

at the end of the 2d year, A

And at the end of the nth year,

P(1+r)2

A = P(1+r)"

This formula may be used in computing the amount of

any sum at compound interest.

1. Find the amount of $200 for 8 years at 5 per cent compound interest.

If the interest is compounded semi-annually, the rate is halved and the number of interest periods doubled; that is

2. Find the principal which will amount to $1000 in 5 years at 6 per cent compound interest.

1. Find the amount of $150 for 7 years at 4 per cent compound interest.

2. Find the amount of $500 for 12 years at 5 per cent compound interest.

3. Of $300 for 9 years 9 months at 5.4 per cent compound interest.

4. Of $600 for 8 years at 6 per cent compounded semiannually.

5. Find the principal that will amount to $500 in 6 years at 4 per cent compound interest.

6. Find the principal that will amount to $1200 in 9 years at 5 per cent compound interest.

7. Find the principal that will amount to $750 in 8 years at 5 per cent compound interest.

8. Find the rate if $648 amounts to $788.38 in 5 years. 9. Find the rate if $9660 amounts to $12945.40 in 6 years. 10. In what time will $4450 amount to $6450 at 31⁄2 per cent compound interest?

24. Given log 2 = 0.3010, find, without table, log 5. 25. When a = 1.05, b = .36, find x,

26. Find the amount of $625.75 for 5 years at 3 per cent compound interest.

27. Find the amount of $1000 for 5 years at 4 per cent compounded semi-annually.

(a + b) 4

=

a1 + 4a3b + 6a2b2 + 4ab3 + b4

To show the law according to which these various terms are formed, these expansions may be written:

Examination shows the following facts:

1. The number of terms is one more than the exponent

of the binomial.

2. The exponent of the first term is equal to the exponent of the binomial.

3. The exponents of a decrease by 1 in successive terms while the exponents of b increase by 1.

4. The sum of the exponents in any term is equal to the exponent of the binomial.

5. The coefficient of the first term is 1, of the second term is equal to the exponent of the first term and thereafter the coefficients are fractions, the numerator consisting of the product of factors decreasing by 1 and the denominator of the same number of factors, starting with 1 and increasing by 1.

If the exponent of the binomial is represented by n, the general form of the expansion is,