(Exponential Equation) How do I find X?

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(Exponential Equation) How do I find X?

$5^{x} = 4^{x + 1}$ $5^x=4^(x+1)$

2 Answers

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Answer 1 · 2017-04-13T13:49:09

$x \approx 6.21$ $x~~ 6.21$ .

Explanation:

Take the natural logarithm of both sides.

$ln (5^{x}) = ln (4^{x + 1})$ $ln(5^x) = ln(4^(x +1))$

Now use ${ln a}^{n} = n ln a$ $lna^n = nlna$ .

$x ln 5 = (x + 1) ln 4$ $xln5 = (x + 1)ln4$

$x ln 5 = x ln 4 + ln 4$ $xln5 = xln4 + ln4$

$x ln 5 - x ln 4 = ln 4$ $xln5 - xln4 = ln4$

$x (ln 5 - ln 4) = ln 4$ $x(ln5 - ln4) = ln4$

This can be simplified further using $ln a - ln b = ln (\frac{a}{b})$ $lna -lnb = ln(a/b)$ .

$x (ln (\frac{5}{4})) = ln 4$ $x(ln(5/4)) = ln4$

$x = \frac{ln 4}{ln (\frac{5}{4})}$ $x = (ln4)/(ln(5/4))$

If you prefer an approximation, we can take $x \approx 6.21$ $x~~ 6.21$ .

Hopefully this helps!

Answer 2 · 2017-04-13T13:50:05

I got: $x = \frac{ln (4)}{ln (5) - ln (4)}$ $x=(ln(4))/(ln(5)-ln(4))$

Explanation:

Here we can try applying the natural log, $ln$ $ln$ , on both sides and apply some properties of logs:
${ln (5)}^{x} = {ln (4)}^{x + 1}$ $ln(5)^x=ln(4)^(x+1)$
then:
$x ln (5) = (x + 1) ln (4)$ $xln(5)=(x+1)ln(4)$
rearrange:
$x ln (5) = x ln (4) + ln (4)$ $xln(5)=xln(4)+ln(4)$
$x ln (5) - x ln (4) = ln (4)$ $xln(5)-xln(4)=ln(4)$
$x [ln (5) - ln (4)] = ln (4)$ $x[ln(5)-ln(4)]=ln(4)$
$x = \frac{ln (4)}{ln (5) - ln (4)}$ $x=(ln(4))/(ln(5)-ln(4))$
if you have a pocket calculator we can easily evaluate the natural log to get:
$x = 6.21256$ $x=6.21256$

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$5^{x} = 4^{x + 1}$ $5^x=4^(x+1)$