How do you solve $2^{x + 3} = 5^{3 x - 1}$ $2^(x+3)=5^(3x-1)$ ?

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How do you solve $2^{x + 3} = 5^{3 x - 1}$ $2^(x+3)=5^(3x-1)$ ?

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Answer 1 · 2016-09-19T02:53:46

$x = \frac{ln 40}{ln 62.5} ≅ 0.892$ $x = ln40/ln62.5 ~= 0.892$

Explanation:

Take the natural logarithm of both sides:

$ln (2^{x + 3}) = ln (5^{3 x - 1})$ $ln(2^(x + 3)) = ln(5^(3x- 1))$

Apply the rule $ln (a^{n}) = n ln a$ $ln(a^n) = nlna$ :

$(x + 3) ln 2 = (3 x - 1) ln 5$ $(x + 3)ln2 = (3x - 1)ln5$

$x ln 2 + 3 ln 2 = 3 x ln 5 - ln 5$ $xln2 + 3ln2 = 3xln5 - ln5$

Isolate the x's to one side:

$3 ln 2 + ln 5 = 3 x ln 5 - x ln 2$ $3ln2 + ln5 = 3xln5 - xln2$

$3 ln 2 + ln 5 = x (3 ln 5 - ln 2)$ $3ln2 + ln5= x(3ln5 - ln2)$

Re-condense using the rule $n ln a = ln (a^{n})$ $nlna = ln(a^n)$ .

$ln 8 + ln 5 = x (ln 125 - ln 2)$ $ln8 + ln5 = x(ln125 - ln2)$

Simplify both sides using the rule $ln a + ln b = ln (a \times b)$ $lna + lnb = ln(a xx b)$ and $ln a - ln b = ln (\frac{a}{b})$ $lna - lnb = ln(a/b)$

$ln (8 \times 5) = x (ln (\frac{125}{2}))$ $ln(8 xx 5) = x(ln(125/2))$

$ln (40) = x (ln (\frac{125}{2}))$ $ln(40) = x(ln(125/2))$

$x = \frac{ln 40}{ln 62.5}$ $x = ln40/ln62.5$

If you want an approximation, you will get $x ≅ 0.892$ $x ~=0.892$ .

Hopefully this helps!

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How do you solve $2^{x + 3} = 5^{3 x - 1}$ $2^(x+3)=5^(3x-1)$ ?

1 Answer

Explanation:

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