Logarithm help needed please

**mysur** · December 3rd 2012, 10:52 AM

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

**MarkFL** · December 3rd 2012, 11:54 AM

a) Assuming we have $a^x=10^{2x+1}$

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

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

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

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

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

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

Solve for $x$ .

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

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

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

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

Equate arguments, and take the valid root.

**Soroban** · December 3rd 2012, 03:53 PM

Hello, mysur!

Are you sure of the instructions?

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

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

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

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

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

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

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

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

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

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

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