Log Help 2 (1 Viewer)

coyazayo · Jan 4, 2016

Hi i got another log questions i need help with. Help will be greatly appreciated. Ty

Q1

ositive integers A,B,and C, with no common factor greater than 1, exists such that
Alog(base200)5+Blog(base200)2=C

What is A+B+C?

Q2: If 11.2^x=1000
0.112^x=?

InteGrand · Jan 4, 2016

coyazayo said:
Hi i got another log questions i need help with. Help will be greatly appreciated. Ty

Q1ositive integers A,B,and C, with no common factor greater than 1, exists such that
Alog(base200)5+Blog(base200)2=C

What is A+B+C?

Q2: If 11.2^x=1000
0.112^x=?

$\underline{QUESTION 1}$

$We have$

$\begin{align*}C&= A \log_{200} 5 + B \log_{200}2 \\ &= \log_{200} \left(5^A\right) + \log_{200} \left(2^B \right) \\ &= \log_{200} \left(5^A\times 2 ^B\right)\\ \Rightarrow 200^C &=5^A\times 2 ^B , \text{but }200 = 2^3 \times 5^2 \text{ so} \\ \left(2^3 \times 5^2 \right)^C &= 5^A\times 2 ^B \\ \Rightarrow 2^{3C} \times 5^{2C} &= 2^B \times 5^A .\end{align*}$

$So $3C=B$ and $2C=A$. Since there are no common factors greater than 1 and all of $A,B,C$ are integers, we set $C=1$, then $B=3$ and $A=2$. So $A+B+C = 1 + 2 + 3 = 6$.$

$\underline{QUESTION 2}$

$We have$

$\begin{align*}11.2^x &= 1000 \\ \Rightarrow x &= \log_{11.2} \left(1000 \right) \\ \Rightarrow \left(0.112\right)^x &= (0.112) ^{\log_{11.2} \left(1000 \right) }\\ &= 1000^{\log_{11.2}(0.112) } \quad (\text{using the log. law }a^{\log_b c}=c^{\log_b a}) \end{align*}$

$and there's really no nice way to simplify this one.$

Log Help 2 (1 Viewer)

coyazayo

Member

InteGrand

Well-Known Member

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