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(log3 2)(log3 4) = log3 8
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A loan of Php 200,000.00 is to be repaid in full after 3 years. If the interest rate is 8% per annum, how much should be paid after 3 years?
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To what factor should you multiply a principal amount if you wish to compute for the interest earned in 6 months of an investment in a fund that pays 24% per annum?
Answer:

Answer 1

Answer:

Step-by-step explanation:

1) The follow logarithmic equations are relevant.

logₐ ᵇ = n

then this implies :

aⁿ = b

(logₐ b × logₐ d) = logₐ b + logₐ d

Applying this we have :

log₃ 2 + log₃ 4 = log₃ 8

Now in finding for n from :

aⁿ = b

Taking log on both sides

log aⁿ = log b

n log a = log b

n = log b/ log a

= log 2/ log3 + log 4/ log 3 = log 8/log 3

0.63093 + 1.2619 = 1.8928

log 8/ log 3 = 1.8928

So it is true they are equal.

2) The compound interest accumulation formula is as follows :

A = p(1 + i)ⁿ

Amount = 200000(1.08)³

= 251942.40

Php 251942.40

3) Since the rate is per annum :

n  = 6/12 = 1/2

i = 24%

The factor should be :

(1 + i)ⁿ

Doing the substitution :

$(1.24)^{1/2}$

Answer 2

"Given:

       $(log^{ 3 }\times 2)(log^{ 3 }\times 4)=log^{ 3 }\times 8$

Solution:

Ans: TRUE

Concept: “$log^n m = n log m$”

3 ($log 2 \times log 4$) = 3 (0.3010 + 0.6020)

                     = 3 (0.903)

3 Log 8 = 3 (0.903)

By equation wise:

        3 (Log 2 \times Log 4) = 3 (Log (2\times4) = 3 (Log 8)

Hence Proved.

Given:

Php = 2,00,000

Term = 3 years

Interest = 8 % per annum = 0.08        

Solution:

Concept: Formula for compound interest is:

A=p{ \left( 1+i \right)}^{ n }

On substitution,

               A = 2,00,000 (1+0.08)^3

                   = 2, 51, 942.40

Given:

Interest = 24% = 0.24

Duration/ term = 6 months = \frac { 1 }{ 2 } year        

Solution:

Concept: Formula for compound interest is:

$A=p{ \left( 1+i \right)}^{ n }$

$A=p{ (1+0.24) }^{ \frac { 1 }{ 2 } }$

            $=p{ (1.24) }^{ \frac { 1 }{ 2 } }$

          A   = 1.1136 p

Where P is the amount to be invested. "