How do you solve $3^{x - 2} \geq 27$ $3^(x-2)>=27$ ?

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Answer 1 · 2016-11-05T11:26:07

$x \geq \frac{27}{ln (3)} + 2$ $x>=27/ln(3)+2$

Note that (using an example) $log (x^{y}) = y log (x)$ $log(x^y) = ylog(x)$

Given $: 3^{x - 2} \geq 27$ $:" "3^(x-2) >=27$

Taking logs - You can use any log. I chose natural logs giving:

$(x - 2) ln (3) \geq 27$ $(x-2)ln(3)>=27$

$x - 2 \geq \frac{27}{ln (3)}$ $x-2>=27/ln(3)$

$x \geq \frac{27}{ln (3)} + 2$ $x>=27/ln(3)+2$

How do you solve 3x−2≥273^(x-2)>=27?