Solving an exponential equation using logarithms

[awelch]

Hi, I could really use some help with this, it's on my test review sheet and I can't figure it out.

10^x + 1
________ = 0.8
10^x + 2


Multiplying both sides by the denominator I get:

10^x + 1 = 0.8 * 10^x + 1.6

but I don't know how to go from that to:

(this step is where I get lost)

10^x = 3

and then,

x = log 3 ≈ 0.477121


If someone could explain how to simplify the original equation into 10^x = 3, I'd really appreciate it. Thanks!

 

[pka]

10^x + 1
________ = 0.8
10^x + 2
Multiplying both sides by the denominator I get:

10^x + 1 = 0.8 * 10^x + 1.6

If someone could explain how to simplify the original equation into 10^x = 3, I'd really appreciate it.

If you had \(\displaystyle y+1=0.8y+1.6\\0.2y=0.6\\y=3\) , then would you be comfortable with that?

If yes, then let \(\displaystyle 10^x=y\),

 

[soroban]

Hello, awelch!

I'll work it out in baby-steps.

\(\displaystyle \dfrac{10^x+1}{10^x+2} \:=\:0.8\)

Multiplying both sides by the denominator I get: .\(\displaystyle 10^x + 1 \:=\: 0.8(10^x) + 1.6\)

We have: .\(\displaystyle 1(10^x) + 1 \:=\:0.8(10^x) +1.6\)

. . \(\displaystyle 1(10^x) - 0.8(10^x) \:=\:1.6 - 1\)

n . . . \(\displaystyle (1-0.8)(10^x) \:=\:0.6\)

. . . . . . . . .\(\displaystyle 0.2(10^x) \:=\:0.6\)

. . . . . . . . . . . . \(\displaystyle 10^x \:=\:\dfrac{0.6}{0.2}\)

. . . . . . . . . . . . \(\displaystyle 10^x \:=\:3\)

. . . . . . . . . . . . . .\(\displaystyle x \:=\:\log3\)

 

[awelch]

Both replies have been very helpful, thank you both a ton!