Question

9^x+5 = 27^x-13
solve using Change of base

Answer 1

please help

Answer 2

\[9^{x+5}=27^{x-13}\]
Am i correct ?

Answer 3

yes

Answer 4

Oh, good catch, the other problem is a bit uglier :-)

Answer 5

\[\left( 3^{2} \right)^{x+5}=\left( 3^{3} \right)^{x-13}\]
can you solve it ?

Answer 6

using the requested change of base, rewrite the equation so that you have the same base on both sides

Answer 7

you distrube?

Answer 8

remember that \[(ab)^n = a^nb^n\]

Answer 9

\[3^2=3*3\]\[3^3=3*3*3\]

Answer 10

\[3^{2x+10}=3^{3x-39}\]

Answer 11

x=49?

Answer 12

Very good!

Answer 13

the "other" equation actually has a solution of \(x=1\) but the other two are pretty ugly, and involve the logs of complex numbers :-)

Answer 14

can you help me @whpalmer4 with that

Answer 15

"other" equation being \[9^x+5 = 27^x-13\]

Answer 16

can you help me with another probkem

Answer 17

problem?

Answer 18

I can try, what's the problem?

Answer 19

solve for x
10^3a-1 = 10^7a+11

Answer 20

there's no \(x\) there :-)

Answer 21

\[10^{3a-1} = 10^{7a+11}\]Solve for \(a\)
Is that the problem?

Answer 22

yes

Answer 23

Okay, as you have two exponentials with the same base and nothing else, isn't it clear that \[3a-1=7a+11\]?

Answer 24

We could take the \(\log_{10}\) of both sides to get that, if you wanted to be formal about it.

Answer 25

ok