How do you solve $5^{x} = 3^{x + 2}$ $5^x = 3^(x+2)$ ?

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Answer 1 · 2018-03-05T18:39:46

$x \approx 4.301 88$ $x~~4.301color(white)(88)$ 3.d.p.

$5^{x} = 3^{x + 2}$ $5^x=3^(x+2)$

Taking logarithms of both sides:

By the law of logarithms:

${ln a}^{b} = b ln (a)$ $lna^b=bln(a)$

$x ln (5) = (x + 2) ln (3)$ $xln(5)=(x+2)ln(3)$

$x ln (5) = x ln (3) + 2 ln (3)$ $xln(5)=xln(3)+2ln(3)$

Subtract $x ln (3)$ $xln(3)$ from both sides:

$x ln (5) - x ln (3) = 2 ln (3)$ $xln(5)-xln(3)=2ln(3)$

Factor $L H S$ $LHS$

$x (ln (5) - ln (3)) = 2 ln (3)$ $x(ln(5)-ln(3))=2ln(3)$

Divide by $(ln (5) - ln (3))$ $(ln(5)-ln(3))$

$x = \frac{2 ln (3)}{ln (5) - ln (3)}$ $x=(2ln(3))/(ln(5)-ln(3))$

$x \approx 4.301 88$ $x~~4.301color(white)(88)$ 3.d.p.

How do you solve 5^x = 3^(x+2)5x=3x+25^x = 3^(x+2)?