Question

What's the real solution for the equation a^x+1= b/a? (consider a>0, a different from 1 and b>0)
a) loga (b)
b) loga (b) - 2
c) loga (b) + 2
d) loga (b+1)

Can someone tell me only how to start solving this?

Answer 1

i take it this is
\[a^{x+1}=\frac{b}{a}\] yes?

Answer 2

That's correct

Answer 3

if so then you can write
\[a\times a^{x+1}=b\]
\[a^{x+2}=b\]
\[x+2=\log_a(b)\]
\[x=\log_a(b)-2\]

Answer 4

a (a^x+1) shouldnt be a^2x+2 ?

Answer 5

@myininaya could you explain that to me pls?

Answer 6

?

Answer 7

I didn't understand the answer. how come a (a^x+1) shouldnt be a^2x+2 ?

Answer 8

We want to show \[a(a^{x+1})=a^{2x+2}?\]

Answer 9

Yes

Answer 10

\[a \cdot a^{x+1}=a^{1} \cdot a^{x+1}=a^{1+(x+1)}=a^{x+1+1}=a^{x+2}\]

Answer 11

Same base and we are multiply add exponents!

Answer 12

is it wrong to do a(a^x+1) => a²^(x+1) and then a^2x+2 ?

Answer 13

I don't understand those steps and yes that is wrong

Answer 14

Thank you so much myini!!