Question

just to double check my answer
@ganeshie8

Answer 1

\[If~\log_ax=1+\log_a(7x-10a)\]express x in terms of a.

Answer 2

i got x=3a/2 and x=-10a^2/(1-7a)

Answer 3

but i think the second answer is wrong

Answer 4

Hmm I understand where the second answer came from...
but where is this 3a/2 coming from?

Answer 5

actually,i hv two workings..i will show the working here..

Answer 6

\[\log_ax=\log_aa+\log_a(7x-10a)\]\[\log_ax-\log_a(7x-10a)=\log_aa\]\[\log_a(x-7x+10a)=\log_aa\]\[-6x=-9a\]\[x=\frac{ 3 }{ 2 }a\]

Answer 7

log(m) - log(n)

is not same as

log(m-n)

Answer 8

\[\log_ax-\log_a(7x-10a)=\log_aa\]\[\log_a \frac{ x }{ (7x-10a) }=\log_aa\]\[\frac{ x }{ 7x-10a }=a\]

Answer 9

\[x=-\frac{ 10a^2 }{ 1-7a }\]

Answer 10

This is my second working..

Answer 11

oh i c.. xD
Thank you @ganeshie8 @zepdrix

Answer 12

yw :)