If $10^{80}=2^x$, what is the value of $x$?

Question

If $10^{80}=2^x$, what is the value of $x$?

(Or, what binary word length would you need to contain $10$ to the $80$?)

Answer 1

Take the log of both sides we will get $80 \log 10=x \log2$, then $x=\frac{80}{\log 2}$. $\log$ is the logarithm of base 10 and $\log 10=1$.

Answer 2

$\ln(b^{x}) = x \times \ln(b)$. Solve for $x$.

Answer 3

We have $10^{80}=2^x.$

Take natural logs of both sides, to give $$\underbrace{\ln[10^{80}]=\ln[2^x] \iff 80\ln(10)=x\ln(2)}_{\text{using the power rule for logarithms}} \iff \boxed{x=\frac{80\ln(10)}{\ln(2)}}=??$$

Alternatively, if $10^{80}=2^x$, then we can immediately deduce that $$\boxed{x=\log_2(10^{80})=80\log_2(10)}=??$$

Answer 4

We have $\frac{1}{\log_{10} 2} \approx 3.3219$. So $2^{3.3219} = 10$, and hence $$10^{80} \approx (2^{3.3219})^{80} = 2^{80\cdot 3.3219} = 2^{265.7542}.$$