• Nov 13th 2007, 08:36 PM
tazrulz99
Hi there,

Can someone please show me how to solve the following equation:

5^x + 25^(2x) = 150

Any help will be greatly appreciated,

Cheers
• Nov 13th 2007, 09:26 PM
janvdl
Quote:

Originally Posted by tazrulz99
Hi there,

Can someone please show me how to solve the following equation:

5^x + 25^(2x) = 150

Any help will be greatly appreciated,

Cheers

Hmm, just check that you didn't make a typo somewhere, it does look a bit like a hard equation for pre-algebra i would say.

\$\displaystyle 5^x + 25^{2x} = 150\$

\$\displaystyle 5^x + 5^{4x} = 150\$

\$\displaystyle Set \ 5^x = k\$

\$\displaystyle k + k^4 = 150\$

\$\displaystyle k^4 + k - 150 = 0\$

I quickly solved it by drawing a graph (See the attached image)

And it seems the answers are -3,54 and 3.46
• Nov 13th 2007, 09:34 PM
tazrulz99
.
Ok and from there you solve with logs. Thanks!
• Nov 13th 2007, 09:46 PM
janvdl
Quote:

Originally Posted by tazrulz99
Ok and from there you solve with logs. Thanks!

Yes because \$\displaystyle k = 5^x\$

But \$\displaystyle 5^x\$ cannot be equal to \$\displaystyle -3,54\$

Thus \$\displaystyle 5^x \$= \$\displaystyle 3,46\$

I found: \$\displaystyle x = 0,77\$