TS EAMCET · Maths · Basic of Mathematics
If \(\log 2=a, \log 3=b, \log 7=c\) and \(6^x=7^{x+4}\) then \(x\) is equal to
- A \(\frac{4 b}{c+a-b}\)
- B \(\frac{4 c}{a+b-c}\)
- C \(\frac{4 c}{c-a-b}\)
- D \(\frac{4 a}{a+b-c}\)
Answer & Solution
Correct Answer
(B) \(\frac{4 c}{a+b-c}\)
Step-by-step Solution
Detailed explanation
We have, \(\begin{array}{cc} & 6^x=7^{x+4} \\ \Rightarrow & x \log 6=(x+4) \log 7 \\ \Rightarrow & x(\log 2+\log 3)=x \log 7+4 \log 7 \\ \Rightarrow & x(\log 2+\log 3-\log 7)=4 \log 7 \\ \Rightarrow & x=\frac{4 \log 7}{\log 2+\log 3-\log 7}=\frac{4 c}{a+b-c} \end{array}\)
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