What is solution of this equation

Question

I am new to logarithms, I am confused what to do when any constant is raised to a power such as $ 1+ \log x $ .

As in this equation,

$$9^{1+\log x} - 3^{1+\log x} -210 = 0$$

While I try to take log on both sides, things get messy.

All the logs have base 3.

score 1 · Accepted Answer · answered Jul 06 '17 at 07:33

1

HINT:

$$9^{1+\log x}=(3^2)^{1+\log x}=(3^{1+\log x})^2$$

Choose $3^{1+\log x}=u$ to form a Quadratic Equation in $u$

$210=14\cdot15$ and for any real $v,3^v>0$

answered Jul 06 '17 at 07:33

lab bhattacharjee

Where does the last line of your answer should be used – Fghj Jul 06 '17 at 07:39
1

@RaviPrakash, We have $$u^2-u-210=(u-15)(u+14)$$ right? So we need to discard $-14$ for real $\log x$ – lab bhattacharjee Jul 06 '17 at 07:40
Exactly, thanks. – Fghj Jul 06 '17 at 07:41

sirous · Answer 2 · 2017-07-07T05:04:34.537

0

$$3^2(1+log x) - 3^(1+log x) =210$$

$$3^(1+log x)[ (3^(1+log x) -1)] = 15 . 14 = 15(15 - 1)$$

$$3^(1+log x) =15 = 3 . 5$$

$$log 3^(1+log x) = 1+log x=log 3 + log 5$$

$$1 + log x = 1 + log 5$$

$$x = 5$$

edited Jul 07 '17 at 05:04

answered Jul 06 '17 at 09:03

sirous

2 Answers2