log help!

[pinkcalculator]

I have to solve this for x, and I'm not quite sure how to start. The brackets around the first term are throwing me off, among other issues.

[log2 (x+3)] ^2- 2log2 (x+3)-15=0.
I think that the bracket placement means I have to solve (log2(x=3))^2 first??
and the second term simplifies to: log2(x+3)^2.
That's about as far as I can get. The (x+3)^2 terms are similar??

Any suggestions would be great, it's really quite frustrating.

 

[fasteddie65]

[log_2 (x + 3)]^2 - 2 log_2 (x + 3) - 15 = 0
[log_2 (x + 3) - 5][log_2 (x + 3) + 3] = 0
log_2 (x + 3) - 5 = 0 or log_2 (x + 3) + 3 = 0
log_2 (x + 3) = 5 or log_2 (x + 3) = -3
(x + 3) = 2^5 = 32 (x + 3) = 2^(-3) = 1/8
x = 29 x = -23/8 (not possible)

 

[soroban]

Hello, pinkcalculator!

If you didn't folow what fasteddie65 did, maybe this will help . . .

\(\displaystyle \text{Solve for x: }\;[\log_2(x+3)] ^2 - 2\log_2 (x+3)-15\:=\:0.\)

\(\displaystyle \text{Replace }\log_2(x+3)\text{ with }u.\)

\(\displaystyle \text{Then we have: }\;u^2 - 2u - 15 \;=\;0\)

See? .It's just a quadratic equation.
And you know how to solve them, right?

\(\displaystyle \text{Factor: }\;(u - 5)(u + 3) \:=\:0 \quad\Rightarrow\quad u \:=\:5,\:-3\)


\(\displaystyle \text{Since }u \,=\,\log_2(x+3)\text{, we have these two equations to solve:}\)

. . \(\displaystyle \log_2(x+3) \:=\:5 \quad\Rightarrow\quad x + 3 \:=\:2^5 \:=\:32 \quad\Rightarrow\quad \boxed{x \:=\:29}\)

. . \(\displaystyle \log_2(x+3) \:=\:-3 \quad\Rightarrow\quad x+3 \:=\:2^{-3} \:=\:\frac{1}{8} \quad\Rightarrow\quad\boxed{ x \:=\:-\frac{23}{8}}\)

And both answers are valid . . .

 

[pinkcalculator]

Thank you so so much!! I can't believe how fast you both responded on a holiday!
I really appreciate the u^2 approach, that made it much more understandable.

Thanks again!