Thread: Using Logarithms to solve for x (form a^x.b^y)

1. Using Logarithms to solve for x (form a^x.b^y)

Hi there, if someone could please help me solve this problem I'd be most grateful:

2^x-3 . 5^x+1 = 62.5

2. Originally Posted by flashylightsmeow
Hi there, if someone could please help me solve this problem I'd be most grateful:

2^x-3 . 5^x+1 = 62.5
I am unsure of how your problem reads, but I think it is this...Please use () in the future.

$2^{x-3}\cdot 5^{x+1}=62.5$ First multiply both sides by $2^4=16$

This gives
$2^{x+1}\cdot 5^{x+1}=1000$ Now since the exponents on the left hand side are the same we can combine the bases to get

$10^{x+1}=1000 \iff 10^{x+1}=10^{3} \implies x+1=3 \implies x=2$

3. AH! Sorry about the layout error. So with problems like these its best to try and make the exponents the same?
Thank you so much!!

Jam