Question

Let logbA=5;logbC=7;logbD=2 what is the value of logb(A^5C^2/D^6)

a. 27

b. there isn't enough info to answer

c. -1

d. 29

Answer 1

To solve for log_b(A^5C^2/D^6), apply the properties of logarithms to simplify the expression, then substitute the known values of log_bA, log_bC, and log_bD. The solution to this problem is 27.

Step-by-step explanation:

The student's question is asking to find the value of logb(A5C2/D6) given that logbA=5, logbC=7, and logbD=2. By using the properties of logarithms, we can break down the expression into separate components before substituting the known values.

Using the product rule for logarithms (log(x*y) = log(x) + log(y)), and the quotient rule (log(x/y) = log(x) - log(y)), we get:

• logb(A5) = 5 * logbA = 5*5
• logb(C2) = 2 * logbC = 2*7
• logb(D6) = 6 * logbD = 6*2

Combining these terms gives us:

logb(A5C2/D6) = logb(A5) + logb(C2) - logb(D6)

Substituting the values, we have:

25 + 14 - 12 = 27

Therefore, the correct answer is 27.

Answer 2

A. 27

Step-by-step explanation:

`\text{De}\text{finition of logarithm}\ \log_ab=c\iff a^c=b.\\\\\log_bA=5\iff A=b^5\\\\\log_bC=7\iff C=b^7\\\\\log_bD=2\iff D=b^2\\\\(A^5C^2)/(D^6)=((b^5)^5(b^7)^2)/((b^2)^6)\qquad\text{use}\ (a^n)^m=a^(nm)\\\\=(b^(25)b^(14))/(b^(12))\qquad\text{use}\ a^n\cdot a^m=a^(n+m)\ \text{and}\ (a^m)/(a^n)=a^(m-n)\\\\=b^(25+14-12)=b^(27)\\\Downarrow\\\\\log_b\left((A^5C^2)/(D^6)\right)=\log_bb^(27)\qquad\text{use}\ \log_aa^n=n\\\\\log_b\left((A^5C^2)/(D^6)\right)=27`