• Dec 3rd 2012, 10:52 AM
mysur
Solve each of the following equation to find x in terms of a where a>0 and a not equal to 100. The logarithms are to base 100
(a) a^x = 10^ 2x+1

(b) 2 log (2x) = 1 + log a
• Dec 3rd 2012, 11:54 AM
MarkFL
a) Assuming we have $\displaystyle a^x=10^{2x+1}$

Take the $\displaystyle \log_{10}$ of both sides:

$\displaystyle \log_{10}(a^x)=\log_{10}(10^{2x+1})$

Apply the property $\displaystyle \log_a(b^c)=c\cdot\log_a(b)$:

$\displaystyle x\cdot\log_{10}(a)=(2x+1)\cdot\log_{10}(10)$

On the right, apply the property $\displaystyle \log_a(a)=1$:

$\displaystyle x\cdot\log_{10}(a)=2x+1$

Solve for $\displaystyle x$.

b) $\displaystyle 2\log_{100}(2x)=1+\log_{100}(a)$

$\displaystyle \log_{100}((2x)^2)=\log_{100}(100)+\log_{100}(a)$

On the right, apply the property $\displaystyle \log_a(b)+\log_a(c)=\log_a(bc)$:

$\displaystyle \log_{100}(4x^2)=\log_{100}(100a)$

Equate arguments, and take the valid root.
• Dec 3rd 2012, 03:53 PM
Soroban
Hello, mysur!

Are you sure of the instructions?

Quote:

Solve each of the following equations to find $\displaystyle x$ in terms of $\displaystyle a$
where $\displaystyle a>0$ and $\displaystyle a \ne 100.$ .The logarithms are to base 100. . Really?

$\displaystyle (b)\;2\log_{100}(2x) \:=\: 1 + \log_{100}(a)$

We have: .$\displaystyle 2\log_{100}(2x) \;=\;1 + \log_{100}(a)$

. . . . . . . . $\displaystyle 2\log_{100}(2x) \;=\;\log_{100}(100) + \log_{100}(a)$

. . . . . . . . $\displaystyle 2\log_{100}(2x) \;=\;\log_{100}(100a)$

. . . . . . . . . $\displaystyle \log_{100}(2x) \;=\;\tfrac{1}{2}\log_{100}(100a)$

. . . . . . . . . $\displaystyle \log_{100}{(2x) \;=\;\log_{100}(100a)^{\frac{1}{2}}$

. . . . . . . . . $\displaystyle \log_{100}(2x) \;=\;\log_{100}(10\sqrt{a})$

I . . . . . . . . . . . . . $\displaystyle 2x \;=\;10\sqrt{a}$

. . . . . . . . . . . . . . . $\displaystyle x \;=\;5\sqrt{a}$