# Thread: Some math problems i cant seem to figure out?

1. ## Some math problems i cant seem to figure out?

9. Solve the equation and express the solution in exact form: log (x+3)=1-logx

10. Solve for indicated variable: P=40,000 X e^t/3 , for t

11. Use any method ( analytic or graphical) to solve the equation. If necessary, round the answer to the nearest thousandth: e^x+ln10=6e^x

2. Originally Posted by rotw121
9. Solve the equation and express the solution in exact form: log (x+3)=1-logx

10. Solve for indicated variable: P=40,000 X e^t/3 , for t

11. Use any method ( analytic or graphical) to solve the equation. If necessary, round the answer to the nearest thousandth: e^x+ln10=6e^x

(9) Assuming that the log is base-10, raise 10 to both sides of the equation to cancel out the logs:
$10^{log(x+3)} = 10^{1-log(x)}$

$x+3 = (10^1)(10^{-log(x)})$

$x+3 = (10)(1/x)$

$x+3 = 10/x$

Can you take it from there?

(b) Take the natural logarithm for both sides:
$P=40000 e^{t/3}$

$ln(P) = ln(40000) + ln(e^{t/3})$

$ln(P) = ln(200^2) + t/3$

$ln(P) = 2 ln(200) + t/3$

Then solve for t from here.

(c) Once again since there is a sort of a $e^x$ or similar expression, take the natural log of both sides
$e^x+ln10=6e^x$

$ln10 = 5e^x$

$\frac{1}{5} ln10 = e^x$

$ln(\frac{1}{5} ln10) = x$

$QED$

3. Originally Posted by rotw121
9. Solve the equation and express the solution in exact form: log (x+3)=1-logx

10. Solve for indicated variable: P=40,000 X e^t/3 , for t

11. Use any method ( analytic or graphical) to solve the equation. If necessary, round the answer to the nearest thousandth: e^x+ln10=6e^x
to #9:

I assume that $\log(x)=\log_{10}(x)$ . The domaion of this equation is $d = \mathbb{R}_0^+$

$\log_{10} (x+3)=1-\log_{10}(x)~\implies~x+3 = \dfrac{10}x$ $~\implies~x^2+3x-10=0~\implies~x=-5~\vee~x=2$

Since $-5\notin d$ the only solution is x = 2

to #10:

$P=40,000 \cdot e^{\frac t3}~\implies~\dfrac P{40000} = e^{\frac t3}~\implies~\dfrac t3=\ln\left(\dfrac P{40000} \right)$

Therefore $t=\ln(P^3)-\ln(40000^3)$

to #11:

If there isn't any typo the result is $x=\ln\left(\dfrac15 \ln(10) \right)$

If and only if you mean:

$e^{x+\ln(10)}=6e^x~\implies~10e^x=6e^x~\implies~10 =6$ which is a rather seldom!