If $2^{n - 7} \times 5^{n - 4} = 1250$ $2^(n-7) \times 5^(n-4) = 1250$ , what is the value of $n$ $n$ ?

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Answer 1 · 2018-07-09T07:48:54

$n = 8$ $n=8$

With exponential equations, one approach is to make the bases on both sides the same.

$2^{n - 7} \times 5^{n - 4} = 1250$ $2^(n-7) xx 5^(n-4) = 1250$

$2^{n - 7} \times 5^{n - 4} = 2 \times 5^{4}$ $2^(n-7) xx 5^(n-4) = 2 xx 5^4$

Now we can compare the indices of like bases.

$n - 7 = 1 \to n = 8$ $n-7 = 1 " " rarr n =8$

$n - 4 = 4 \to n = 8$ $n-4 = 4" "rarr n = 8$

Do not be tempted to multiply the $2 and 5!$ $2 and 5!$

The indices are different so you may not.

If 2n−7×5n−4=1250 2^(n-7) \times 5^(n-4) = 1250, what is the value of nn?