How do you solve $9^{5 k} + 8 = 50$ $9^(5k)+8=50$ ?

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Answer 1 · 2016-10-11T09:22:36

$k = 0.3402$ $k=0.3402$

As $9^{5 k} + 8 = 50$ $9^(5k)+8=50$ ,

$9^{5 k} = 50 - 8 = 42$ $9^(5k)=50-8=42$

or taking logarithm to base $10$ $10$ , we get

$5 k log 9 = log 42$ $5klog9=log42$

or $k = \frac{log 42}{5 log 9}$ $k=log42/(5log9)$

or $k = \frac{1.62325}{5 \times 0.95424}$ $k=1.62325/(5xx0.95424)$

or $k = 0.3402$ $k=0.3402$

How do you solve 95k+8=509^(5k)+8=50?