Question

5(1.06^2x+1)=11

Answer 1

What's the question?\[5(1.06^2x+1)=11?\]\[5(1.06^{2x}+1)=11?\]\[5(1.06^{2x+1})=11?\]

Answer 2

use logarithms to solve (round 4 decimal places) sorry.

Answer 3

:)

I meant, which of the three I posted above looks like what you've been asked to solve?

Answer 4

oh and it's the third one 5(1.062x+1)=11

Answer 5

yes. i'm sorry

Answer 6

the third one you wrote ignore everything else.

Answer 7

The first thing you should do is get rid of that five like this\[1.06^{2x+1}=\frac{11}{5}.\]Now you have to take the logarithm with base 1.06 of both sides:\[\log_{1.06}1.06^{2x+1}=log_{1.06}\frac{11}{5}\implies2x+1=log_{1.06}\frac{11}{5}.\]Now you solve for x.

Answer 8

that
equaled 13.5313 and the book says it should equal 6.2657

Answer 9

\[x=\frac{1}{2}\log_{1.06}\frac{11}{5}-\frac{1}{2}\approx6.2657\]

Answer 10

So it's right. :)

Answer 11

ok thanks. we just did it in class...
x= (log(11/5)/log(1.06) divided by 2 = 6.765678179 thank you