I need help in this calculus problem.?

Question

solve for x,

5^2x=2^5x+1

im stuck

i took LN(natural log) of both sides, but im not sure what to do next.

please help due 2marrow

thanks!

Answer 1

If the numbers I have parenthesized in the top line below are supposed to be exponents, then here is your solution.  After taking the logs, multiply the equation out as though you were solving a linear equation and then solve for x:

5^(2x) = 2^(5x + 1)

(2x) ln 5 = (5x + 1) ln 2

(2x)(1.60944) ≈ (5x + 1)(0.69315)

3.21888x ≈ 3.46575x + 0.69315

-0.69315 ≈ 3.46575x - 3.21888x

-0.69315 ≈ 0.24687x

-0.69315 / 0.24687 ≈ x

-2.80775 = x

To verify, just plug the calculated value for x into the original equation.  Assuming that the parenthesized numbers are exponents, then the equation balances down to the fourth decimal place.

5^(2x) = 2^(5x + 1)

5^(-5.6155) ≈ 2^(-13.0388)

1.18832 x 10^-4 ≈ 1.18831 x 10^-4

Since the equation balances so closely, we have approximately the correct value for x.

x ≈ -2.80775.

Notice that if you use Catalina's solution below, approximately the same numerical value for x is found.  The two values agree down to the third decimal place:

x = ln 2 / ln (25/32)

x ≈ 0.69315 / -0.24686

x ≈ -2.80786

Answer 2

ln(5^2x) = ln (2^5x+1)

2xln(5)=(5x+1)ln(2)

2xln(5)=5xln(2)+ln(2)

2xln(5) - 5xln(2) = ln(2)

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

x = ln(2)/(2ln(5) - 5ln(2))... I don't know how far you want to simplify

ln(25) - ln(32)=

x= ln(2)/ln(25/32)