# Logarithm

• Jul 18th 2009, 01:07 AM
Rose Wanjohi
Logarithm
solve the equation
(2^x)(3^(x+1)=10
• Jul 18th 2009, 01:13 AM
e^(i*pi)
Quote:

Originally Posted by Rose Wanjohi
solve the equation
(2^x)(3^(x+1)=10

Assuming the question is $\displaystyle (2^x)(3^{x+1}) = 10$

Hint: $\displaystyle 3^{x+1} = 3 \times 3^x$

I get an answer of 0.672 (3sf).

I've put my working in a spoiler below so you can check your own too

Spoiler:
$\displaystyle (3)(2^x)(3^x) = 10$

Take the logs (base 10 will keep it simpler) $\displaystyle log(3 \times 2^x \times 3^x) = 1$

Use the law $\displaystyle log_b(ac) = log_b(a) + log_b(c)$ to separate the left hand side

$\displaystyle log(3) + log(2^x) + log(3^x) = 1$

Take log(3) from both sides and then use the law which says $\displaystyle log(a^k) = k\, log(a)$

$\displaystyle x\, log(2) + x\, log(3) = 1-log(3)$

Factor out x: $\displaystyle x(log(2)+log(3)) = 1-log(3)$

Divide through to isolate x: $\displaystyle \frac{1-log(3)}{log(2)+log(3)}$
• Jul 18th 2009, 01:24 AM
Rose Wanjohi
logarithm
yeah i got it,
xlog2+xlog3+log3=1
log3=0.4771
log2=0.3010
therefore 0.3010x+0.4771x+0.4771=1