Question

Please solve and don't spam or write anything unnecessarily it is a request ,question given below .
$\frac{ log(a) }{5} = \frac{ log(b) }{6} = \frac{ log(c) }{7} . \: then \: {b}^{2} = \: ?$
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Answer 1

$\LARGE{ \underline{ \purple{ \sf{Required \: answer:}}}}$

GiveN:

• $\rm{ \dfrac{ log(a) }{5} = \dfrac{ log(b) }{6} = \dfrac{ log(c) }{7} }$

We have to find the value of b² according to the above data.

Let,

$\rm{ \dfrac{ log(a) }{5} = \dfrac{ log(b) }{6} = \dfrac{ log(c) }{7} = k}$

Then,

• log a = 5k
• log b = 6k
• log c = 7k

When the base of the logarithm is not mentioned, it is taken to be of base 10.

• a = $\rm{{10}^{5k} }$
• b = $\rm{{10}^{6k} }$
• c = $\rm{{10}^{5k} }$

We have to find the value of b²

⇛ $\rm{({{10}^{6k}})^{2} }$

⇛ $\rm{{10}^{12k} }$

This can be written as,

⇛ $\rm{{10}^{5k + 7k} }$

⇛ $\rm{{10}^{5k} }\times\rm{{10}^{7k} }$

⇛ a × c

Thus, the required value of b² is ac.

And we are done...

Answer 2

$\huge\mathbb\red{Answer:-}$

GiveN:

$\rm{ \dfrac{ log(a) }{5} = \dfrac{ log(b) }{6} = \dfrac{ log(c) }{7} }5log(a)=6log(b)=7log(c)$

We have to find the value of b² according to the above data.

Let,

$\rm{ \dfrac{ log(a) }{5} = \dfrac{ log(b) }{6} = \dfrac{ log(c) }{7} = k}5log(a)=6log(b)=7log(c)=k$

Then,

log a = 5klog b = 6klog c = 7k

When the base of the logarithm is not mentioned, it is taken to be of base 10.

$a = \rm{{10}^{5k} }105kb = \rm{{10}^{6k} }106kc = \rm{{10}^{5k} }105k$

We have to find the value of b²

⇛ $\rm{({{10}^{6k}})^{2} }(106k)2$

⇛ $\rm{{10}^{12k} }1012k$

This can be written as,

⇛ $\rm{{10}^{5k + 7k} }105k+7k$

⇛ $\rm{{10}^{5k} }\times\rm{{10}^{7k} }105k×107k$

⇛ a × c

Thus, the required value of b² is ac.

And we are done...