How do you solve for x in $2^{x} = 5^{x + 6}$ $2^x = 5^(x+6)$ ?

Question

How do you solve for x in $2^{x} = 5^{x + 6}$ $2^x = 5^(x+6)$ ?

1 Answer

Impact of this question

1258 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-20T20:57:23

Taking the logarithm on both sides we get
$x ln (2) = (x + 6) ln (5)$ $xln(2)=(x+6)ln(5)$
rearranging we obtain
$x (ln (2) - ln (5)) = 6 ln (5)$ $x(ln(2)-ln(5))=6ln(5)$
so
$x = \frac{6 ln (5)}{ln (2) - ln (5)}$ $x=(6ln(5))/(ln(2)-ln(5))$

Explanation:

We used that $ln (x^{r}) = r ln (x)$ $ln(x^r)=rln(x)$

Science

Math

Humanities

... and beyond

How do you solve for x in $2^{x} = 5^{x + 6}$ $2^x = 5^(x+6)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve for x in 2^x = 5^(x+6)2x=5x+62^x = 5^(x+6)?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve for x in $2^{x} = 5^{x + 6}$ $2^x = 5^(x+6)$ ?