If log 7 x-log 13 x-1 and x-13log x 7- then k is divisible by

The correct options are A 7 B 13Given : nbsp;log _7x + log_13x = 1 and x = 13^log_k7 Now, nbsp; x = 13^log_k7 ⇒log_13 x =log_k 7 x = 7^log_k 13 ⇒log_7 x =l ...

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Byju's Answer

Standard XII

Mathematics

Fundamental Laws of Logarithms

If log 7 x+lo...

Question

If ${log}_{7} x + {log}_{13} x = 1$ and $x = 13^{{log}_{k} 7}$ , then $k$ is divisible by

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Solution

The correct options are
A $7$
B $13$
Given : ${log}_{7} x + {log}_{13} x = 1$ and $x = 13^{{log}_{k} 7}$
Now,
$x = 13^{{log}_{k} 7} \Rightarrow {log}_{13} x = {log}_{k} 7 x = 7^{{log}_{k} 13} \Rightarrow {log}_{7} x = {log}_{k} 13$
Now,
${log}_{7} x + {log}_{13} x = 1 \Rightarrow {log}_{k} 7 + {log}_{k} 13 = 1 \Rightarrow {log}_{k} 91 = 1 ∴ k = 91$

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If log7x+log13x=1 and x=13logk7, then k is divisible by

The correct options are A 7 B 13Given : log7x+log13x=1 and x=13logk7 Now, x=13logk7⇒log13x=logk7x=7logk13⇒log7x=logk13 Now, log7x+log13x=1⇒logk7+logk13=1⇒logk91=1∴k=91

If ${log}_{7} x + {log}_{13} x = 1$ and $x = 13^{{log}_{k} 7}$ , then $k$ is divisible by