# Can it be simplified more?

• Jul 29th 2010, 06:06 AM
danielh9103
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?
• Jul 29th 2010, 06:16 AM
Also sprach Zarathustra
ln(20)=ln(4*5)=ln(4)+ln(5)=ln(2^2)+ln(5)=2ln2+ln(5 )
• Jul 29th 2010, 06:21 AM
eumyang
Quote:

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
$5^{3 + x} = 20^{x - 3}$

So,
\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...
\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}