Heidi Tate

2023-03-12

If ${10}^{x}=64$ then what is the value of ${10}^{(x/2)}+1$

fodheargnl0

Beginner2023-03-13Added 4 answers

We have ${10}^{x}=64$.

Now

$={10}^{x}/2+1\phantom{\rule{0ex}{0ex}}={10}^{x}/2\times {10}^{1}\phantom{\rule{0ex}{0ex}}=({10}^{x}{)}^{1}/2\times {10}^{1}\phantom{\rule{0ex}{0ex}}=(64{)}^{1}/2\times 10\phantom{\rule{0ex}{0ex}}=8\times 10\phantom{\rule{0ex}{0ex}}({64}^{1/2}=squareroot(64)=8)$

= 80

Now

$={10}^{x}/2+1\phantom{\rule{0ex}{0ex}}={10}^{x}/2\times {10}^{1}\phantom{\rule{0ex}{0ex}}=({10}^{x}{)}^{1}/2\times {10}^{1}\phantom{\rule{0ex}{0ex}}=(64{)}^{1}/2\times 10\phantom{\rule{0ex}{0ex}}=8\times 10\phantom{\rule{0ex}{0ex}}({64}^{1/2}=squareroot(64)=8)$

= 80