Question

please solve it ...........................​

Answer 1

Answer:

$\implies n = \frac{7log2+13log5}{log2+log5}$

Step-by-step explanation:

$2^{n-7}\times 5^{n-4}=5^{9}$

$\implies 2^{n-7}=\frac{5^{9}}{5^{n-4}}$

$\implies 2^{n-7}=5^{9-(n-4)}$

$\implies 2^{n-7}=5^{9-n+4)}\\=5^{13-n}$

/* Take logarithm both sides of the equation, we get

$\implies log2^{n-7}=log5^{13-n}$

$\implies (n-7)log2=(13-n)log5$

$\implies nlog2 -7log2=13log5-nlog5$

$\implies nlog2 +nlog 5=7log2+13log5$

$\implies n(log2+log5)=7log2+13log5$

$\implies n = \frac{7log2+13log5}{log2+log5}$

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