How do you solve $2 {(1 + 0.1)}^{x} = 50$ $2(1+0.1)^x=50$ ?

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How do you solve $2 {(1 + 0.1)}^{x} = 50$ $2(1+0.1)^x=50$ ?

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Answer 1 · 2017-01-22T20:19:33

$x \approx 33.77 to 2 dec. places$ $x≈33.77" to 2 dec. places"$

Explanation:

Isolate the exponential by dividing both sides by 2

$\Rightarrow {(1.1)}^{x} = 25$ $rArr(1.1)^x=25$

$color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(log_b x=nhArrx=b^n)color(white)(2/2)|)))$

Take ln ( natural log) of both sides.

$rArrln(1.1)^x=ln25$

$rArrxln(1.1)=ln25$

$rArrx=(ln25)/(ln1.1)≈33.77" to 2 decimal places"$ #

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How do you solve $2 {(1 + 0.1)}^{x} = 50$ $2(1+0.1)^x=50$ ?

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Explanation:

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