# Logarithmic/Exponential Equation

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

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

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

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

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

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

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

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

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

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

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

$5x = 7 + 6$

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

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

Quote:

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

Work:

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

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

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

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

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

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

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

Your second method was faulty . . .

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

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

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