Please explain in detail why this problem should be calculated in this way? In order to buy a house, Mr. Zhang borrows from the bank. The amount of loan is d yuan. He is prepared to repay P yuan every month. The monthly interest rate is r M=logP-log(P-DxR)/log(1+R) On the first floor, I don't understand how you calculate these two formulas: the sum of principal and interest of repayment is p (1 + R) ^ (m-1) + P (1 + R) ^ (m-2) +... + P = P [(1 + R) ^ M-1] / R, Let D (1 + R) ^ m = P [(1 + R) ^ M-1] / R, We get (1 + R) ^ m = P / (P-D * r),

Please explain in detail why this problem should be calculated in this way? In order to buy a house, Mr. Zhang borrows from the bank. The amount of loan is d yuan. He is prepared to repay P yuan every month. The monthly interest rate is r M=logP-log(P-DxR)/log(1+R) On the first floor, I don't understand how you calculate these two formulas: the sum of principal and interest of repayment is p (1 + R) ^ (m-1) + P (1 + R) ^ (m-2) +... + P = P [(1 + R) ^ M-1] / R, Let D (1 + R) ^ m = P [(1 + R) ^ M-1] / R, We get (1 + R) ^ m = P / (P-D * r),

First of all, we need to understand what repayment is. When repayment is completed, the sum of principal and interest of the loan should be equal to the sum of principal and interest of repayment,
Suppose that the payment is made in m months (and the repayment time is at the end of the month),
Then the sum of principal and interest is d (1 + R) ^ M,
The sum of principal and interest of repayment is p (1 + R) ^ (m-1) + P (1 + R) ^ (m-2) +... + P = P [(1 + R) ^ M-1] / R,
Let D (1 + R) ^ m = P [(1 + R) ^ M-1] / R,
We get (1 + R) ^ m = P / (P-D * r),
Take the logarithm Mlog (1 + R) = logP log (p-dr) on both sides,
M = [logP log (P-D * r)] / log (1 + R)

How to multiply the same base logarithm

log(a)(b)=1/log(b)(a)log(a)(N)=log(b)(N)÷log(b)(a)1、a^(log(a)(b))=b 2、log(a)(a^b)=b 3、log(a)(MN)=log(a)(M) log(a)(N); 4、log(a)(M÷N)=log(a)(M)-log(a)(N); 5、log(a)(M^n)=nlog(a)(M) 6、log(a^n)M=1/...

1、 (1) according to Logan = bab = n, the formula of changing bottom is proved
LogaN=LogmN/Logma
(2) Use the bottom changing formula in (1) to find the value of the following formula
Log225*Log34*Log59 ；
(3) Using the bottom changing formula in (1) to prove
Logab*Logbc*Logca=1 .
2、 Let f (x) = 3x;
（1）f(x)*f(y)=f(x+y) ;
（2）f(x)/f(y)=f(x-y)
3、 Let f (x) = LG (1-x) / (1 + x), a, B ∈ (- 1,1), and prove it
f(a)+f(b)=f[(a+b)/(1+ab)]

（1）
Let a ^ B = n ①
Then B = Logan ②
The logarithm identity is obtained by substituting (2) into (1)
a^(logaN)=N………… ③
Take the logarithm of m as the base on both sides of 3
logaN·logma=logmN
therefore
logaN=(logmN)/(logma）
（2）
Original formula = 2log2 (5) · 2log3 (2) · 2log5 (3)
=8（lg5/lg2)(lg2/lg3)(lg3/lg5)
=8
(3)
The original formula = (LGB / LGA) (LGC / LGB) (LGA / LGC) = 1
The meaning of LG (x) above is clear, with 10 as the base