LOG QUESTIONs please help

**pal4499** · November 4th 2008, 09:53 AM

need some help with these questions PLEASE

1.
solve the equation

0.5 ^(1+3x)=7

giging you answer to 2 decimal places

2.
log (5x) - log(1/x+1)=2log(2x+2)

**earboth** · November 4th 2008, 10:26 AM

Originally Posted by pal4499

need some help with these questions PLEASE

1.
solve the equation

0.5 ^(1+3x)=7

giging you answer to 2 decimal places

2.
log (5x) - log(1/x+1)=2log(2x+2)

to #1:

$\left(\dfrac12\right)^{1+3x} = 7~\iff~\dfrac12 \left(\dfrac18\right)^x=7~\implies~$ $\left(\dfrac18\right)^x=14~\implies~8^x=\dfrac1{14 }~\implies~x=\log_8\left(\dfrac1{14}\right)~\impli es~$ $x=\log_8\left(14^{-1}\right)~\implies~x=\dfrac{-\ln(14)}{\ln(8)}$

to #2:

Unfortunately the writing isn't clear. I assume that you mean:

$\log(5x)-\log\left(\dfrac1{x+1}\right)=2\log(2x+2)~\implies ~\log(5x(x+1))=\log((2x+2)^2)$

Solve for x:

$5x(x+1)=(2x+2)^2~\iff~5x(x+1)=4(x+1)^2~\implies~x=-1~\vee~x=4$

Keep in mind that the log-function is defined for positive values only and therefore x = -1 is not a solution of this equation!

EDIT: Removed a mistake

**pal4499** · November 4th 2008, 10:45 AM

i may have written the question out wrong!!

(1)
0.5(1+3x)=7

**earboth** · November 4th 2008, 11:09 AM

Originally Posted by pal4499

i may have written the question out wrong!!

(1)
0.5(1+3x)=7

Remember that .... $0.5=\dfrac12$
So I've done your question in my previous post. I didn't give you the approximative number because I was pretty sure that you can calculate it yourself. For confirmation $x\approx -1.27$

EDIT: I've corrected my previous post. Have a look!

Thread: LOG QUESTIONs please help

LinkBack

Thread Tools

Display

LOG QUESTIONs please help

Similar Math Help Forum Discussions

More log questions

Please someone help me with just 2 questions?

Some Questions !

Plz solve these questions.(Trignometry statement questions)

Assorted algebraic questions ( Read this before asking some questions! )

Search Tags