# Thread: Can it be simplified more?

1. ## Can it be simplified more?

Hi

Solving leaving the answer in term of e or ln

(5^3+x) = (20^x-3)

ln 5 (3 + x) = ln 20 (x - 3)

3 ln5 + x ln5 = x ln20 - 3 ln20

3 ln5 + 3 ln20 = x(ln20 - ln5)

can it be simplified more?

2. ln(20)=ln(4*5)=ln(4)+ln(5)=ln(2^2)+ln(5)=2ln2+ln(5 )

3. Originally Posted by danielh9103
Hi

Solving leaving the answer in term of e or ln

(5^3+x) = (20^x-3)

ln 5 (3 + x) = ln 20 (x - 3)

3 ln5 + x ln5 = x ln20 - 3 ln20

3 ln5 + 3 ln20 = x(ln20 - ln5)

can it be simplified more?
You should start learning LaTeX - this is hard to read. I assume that you mean
$\displaystyle 5^{3 + x} = 20^{x - 3}$

So,
\displaystyle \begin{aligned} \ln 5^{3 + x} &= \ln 20^{x - 3} \\ 3 \ln 5 + x \ln 5 &= x \ln 20 - 3 \ln 20 \\ 3 \ln 5 + 3 \ln 20 &= x(\ln 20 - \ln 5) \end{aligned}

This is what you wrote so far. But you're not done yet...
\displaystyle \begin{aligned} 3 (\ln 5 + \ln 20) &= x(\ln 20 - \ln 5) \\ 3 \ln 100 &= x \ln 4 \\ x &= \frac{3 \ln 100}{\ln 4} \\ x &= \frac{6 \ln 10}{\ln 4} \\ \end{aligned}