Dear Student , Please find below the solution to the asked query : 49A+5B=491-log72 + 5-log54=4949log72+15log54=492log749+14log55 Since mlogax=xlogam=492log772+14 Since logaa=1=4922 log77+14=494+14=504=252=12 5 ANS Hope this information will clear your doubts about the topic If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible Regards

solve que 6

Dear Student ,

Please find below the solution to the asked query :

49^{A} + 5^{B} = 49^{1 - \log_{7} 2} + 5^{- \log_{5} 4} = \frac{49}{49^{\log_{7} 2}} + \frac{1}{5^{\log_{5} 4}} = \frac{49}{2^{\log_{7} 49}} + \frac{1}{4^{\log_{5} 5}} (Since m^{\log_{a} x} = x^{\log_{a} m}) = \frac{49}{2^{\log_{7} 7^{2}}} + \frac{1}{4} (Since \log_{a} a = 1) = \frac{49}{2^{2 \log_{7} 7}} + \frac{1}{4} = \frac{49}{4} + \frac{1}{4} = \frac{50}{4} = \frac{25}{2} = 12.5 ANS . . .

Hope this information will clear your doubts about the topic .

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible .

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solve que 6​

solve que 6