Logarithms

[horsewoman27]

Solve for x.

log3(x+7)=2-log3(x-1)

Write the exact answer using base-10 logarithms.


log3(x+7)=2-log3(x-1)
log3(x+7) + log3(x-1) = 2
log3 [(x+7)/(x-1)] = 2
(x+7)(x-1) = 3^2
x^2 + 6x - 7 = 9
x^2 + 6x - 16 = 0
(x+8)(x-2) = 0
x = -8 or x = 2

but x=-8 would make log3(x+7) undefined, so

x = 2

( why would it ask to use base 10 logs ????)

 

[Deleted member 4993]

Solve for x.

log[sub:3f7qy7pk]3[/sub:3f7qy7pk](x+7)=2-log[sub:3f7qy7pk]3[/sub:3f7qy7pk](x-1)

Write the exact answer using base-10 logarithms.


log[sub:3f7qy7pk]3[/sub:3f7qy7pk](x+7)=2-log[sub:3f7qy7pk]3[/sub:3f7qy7pk](x-1)
log[sub:3f7qy7pk]3[/sub:3f7qy7pk](x+7) + log[sub:3f7qy7pk]3[/sub:3f7qy7pk](x-1) = 2
log[sub:3f7qy7pk]3[/sub:3f7qy7pk] [(x+7)*(x-1)] = 2 <<<< That should be multiplication.
(x+7)(x-1) = 3^2
x^2 + 6x - 7 = 9
x^2 + 6x - 16 = 0
(x+8)(x-2) = 0
x = -8 or x = 2

but x=-8 would make log[sub:3f7qy7pk]3[/sub:3f7qy7pk](x+7) undefined, so

x = 2

( why would it ask to use base 10 logs ????)

Good work - and thank you for showing it.

May be the question meant to ask you to show answer in base-10 (decimal system) numbers - not log[sub:3f7qy7pk]10[/sub:3f7qy7pk]
.

 

[horsewoman27]

I guess I still don't understand how to answer in that format

 

[Deleted member 4993]

I guess I still don't understand how to answer in that format

x = 2 is the answer in decimal format