# Logarithmic/Exponential Equation

• Dec 13th 2007, 03:09 PM
Macleef
Logarithmic/Exponential Equation
What the heck did I do wrong in the following problem?

Solve: $\displaystyle 2^{5x-6} = 7$

Work:
$\displaystyle log2^{5x-6} = log7$

$\displaystyle (5x - 6) (log2) = log7$

$\displaystyle 5x - 6 = (\frac {log7}{log2})$

$\displaystyle 5x = (\frac {log7}{log2}) + 6$

$\displaystyle x = \frac {[(\frac {log7}{log2}) + 6]}{5}$

And when I put both sides to base of 2, I get:

$\displaystyle log_22^{5x-6} = log_22^{7}$

$\displaystyle (5x-6)(log_22) = 7(log_22)$

$\displaystyle (5x-6) = \frac {7(log_22)}{(log_22)}$

$\displaystyle 5x = 7 + 6$

$\displaystyle x = \frac{7+6}{5}$

Textbook Answer: $\displaystyle \frac {log_27 + 6}{5}$
• Dec 13th 2007, 03:32 PM
Soroban
Hello, Macleef!

Quote:

Solve: .$\displaystyle 2^{5x-6} = 7$

Work:

$\displaystyle \log2^{5x-6} \:= \:\log7$

$\displaystyle (5x - 6)\log2) \:= \:\log7$

$\displaystyle 5x - 6 \:= \:\frac{\log7}{\log2}$

$\displaystyle 5x \:= \:\frac{\log7}{\log2} + 6$

$\displaystyle x \:= \:\frac{\dfrac{\log7}{\log2} + 6}{5}$ . . . . Right!

Textbook Answer: $\displaystyle \frac{\log_2\!7 + 6}{5}$ . . . . also correct!

Using the Base-Change Formula, note that: .$\displaystyle \log_2\!7 \:=\:\frac{\log 7}{\log 2}$

Your second method was faulty . . .

Take logs, base 2: .$\displaystyle \log_2\left(2^{5x-6}\right) \:=\:\log_2(7)$

$\displaystyle \text{Then we have: }\;(5x-6)\underbrace{\log_2(2)}_{\text{This is 1}} \:=\:\log_2(7)$

Therefore: .$\displaystyle 5x - 6 \:=\:\log_2(7) \quad\Rightarrow\quad x \:=\:\frac{\log_2(7) + 6}{5}$